# Riemann sums (3)

Problem. Given integer $k \ge 2,$ show that

$\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^n \left(\left \lfloor \frac{kn}{j} \right \rfloor - k \left \lfloor \frac{n}{j} \right \rfloor \right)=k(\ln k +1-H_k),$

where $\lfloor \ \ \rfloor$ is the floor function and $\displaystyle H_k=\sum_{i=1}^k \frac{1}{i}$ is the $k$-th harmonic number.

Solution. We have

\displaystyle \begin{aligned} I:=\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^n \left(\left \lfloor \frac{kn}{j} \right \rfloor - k \left \lfloor \frac{n}{j} \right \rfloor \right)=\int_0^1 \left (\left \lfloor \frac{k}{x} \right \rfloor - k \left \lfloor \frac{1}{x} \right\rfloor \right) \ dx \\ =\int_1^{\infty} \frac{\lfloor kx \rfloor - k\lfloor x \rfloor}{x^2} \ dx.\end{aligned} \ \ \ \ \ \ \ \ (*)

Now see that, for integers $m,$ we have $\lfloor kx \rfloor - k\lfloor x \rfloor=i$ if

$\displaystyle m+\frac{i}{k} \le x < m+\frac{i+1}{k}, \ \ i=0, 1, \cdots , k-1.$

Thus, by $(*),$

\displaystyle \begin{aligned} I=\sum_{m=1}^{\infty} \sum_{i=1}^{k-1} \int_{m+i/k}^{m+(i+1)/k} \frac{i}{x^2} \ dx =k\sum_{m=1}^{\infty} \sum_{i=1}^{k-1}i\left(\frac{1}{km+i}-\frac{1} {km+i+1}\right)=k\sum_{m=1}^{\infty}\left(\sum_{i=1}^k\frac{1}{km+i}-\frac{1}{m+1}\right) \\ =k \lim_{N\to\infty} \sum_{m=1}^{N-1}\left(\sum_{i=1}^k\frac{1}{km+i}-\frac{1}{m+1}\right)=k \lim_{N\to\infty}\left(\sum_{i=k+1}^{Nk} \frac{1}{i}-\sum_{i=2}^N \frac{1}{i}\right)=k\lim_{N\to\infty}(H_{Nk}-H_k-H_N+1) \\ =k \lim_{N\to\infty}(H_{Nk}-H_N)+k(1-H_k) =k\ln k+k(1-H_k)=k(\ln k +1-H_k). \ \Box\end{aligned}

Exercise. In the solution of the above problem, I used the following identities

i) $\displaystyle \sum_{i=1}^{k-1}i\left(\frac{1}{km+i}-\frac{1} {km+i+1}\right)=\sum_{i=1}^k\frac{1}{km+i}-\frac{1}{m+1}$

ii) $\displaystyle \lim_{N\to\infty}(H_{Nk}-H_N)=\ln k.$

Explain why they are true!
Hint. For ii), see Exercise 1 in this post for a big hint!

# Harmonic numbers (3)

Notation. Throughout this post, $\displaystyle a_n:=\frac{\binom{2n}{n}}{4^n},$ where $n \ge 0$ is an integer.

We have already seen the first part of the following problem a couple of times in this blog (see Exercise 5 in here and here!).

Problem 1. Show that

i) $\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{n}x^n=2\ln 2 - 2 \ln(\sqrt{1-x}+1),$ for $x \in [-1,1].$ In particular, $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n}=2\ln 2$

ii) $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n^2}=\frac{\pi^2}{6}-2(\ln 2)^2.$

Solution. i) Since $\displaystyle a_n=(-1)^n\binom{-1/2}{n},$ (see the Example in this post!), we have $\displaystyle \sum_{n=0}^{\infty}a_nx^n=\frac{1}{\sqrt{1-x}},$ for $x \in (-1,1).$ Thus $\displaystyle \sum_{n=1}^{\infty}a_nx^{n-1}=\frac{1}{x\sqrt{1-x}}-\frac{1}{x}$ and hence

$\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{n}x^n=\int_0^x\left(\frac{1}{t\sqrt{1-t}}-\frac{1}{t}\right)dt=2\ln 2 - 2 \ln(\sqrt{1-x}+1). \ \ \ \ \ \ \ \ (*)$

Now since the interval of convergence of $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n}x^n$ is $[-1,1],$ (why ?), we may put $x=1$ in $(*),$ and get $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n}=2\ln 2.$

ii) By i), we have

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{a_n}{n^2}=\sum_{n=1}^{\infty} \frac{a_n}{n} \int_0^1 x^{n-1} dx =\int_0^1 \sum_{n=1}^{\infty} \frac{a_n}{n}x^{n-1} dx=\int_0^1 \frac{2\ln 2 - 2 \ln(\sqrt{1-x}+1)}{x} \ dx \\ =-2\int_0^1\frac{1}{x}\ln\left(\frac{\sqrt{1-x}+1}{2}\right) dx.\end{aligned}

So the substitution $\displaystyle \frac{\sqrt{1-x}+1}{2}=t$ gives

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{a_n}{n^2}=2\int_{1/2}^1 \frac{1-2t}{t(1-t)} \ln t \ dt=2\left(\int_{1/2}^1 \frac{\ln t}{t} \ dt - \int_{1/2}^1 \frac{\ln t}{1-t} \ dt\right)=-(\ln 2)^2-2\int_{1/2}^1 \frac{\ln t}{1-t} \ dt \\ =-(\ln 2)^2 - 2\int_0^{1/2} \frac{\ln(1-t)}{t} \ dt=-(\ln 2)^2+2\text{Li}_2(1/2)=-2(\ln 2)^2+\frac{\pi^2}{6}, \end{aligned}

by the remark and Problem 1 in this post. $\Box$

Problem 2. Show that $\displaystyle \sum_{n=1}^{\infty} \frac{H_na_n}{n}=\frac{\pi^2}{3}.$

Solution. We have

\displaystyle \begin{aligned} \sum_{n=1}^{\infty}\frac{H_na_n}{n}=\sum_{n=1}^{\infty}\frac{a_n}{n}\int_0^1\frac{1-x^n}{1-x} \ dx=\int_0^1\frac{1}{1-x}\left(\sum_{n=1}^{\infty}\frac{a_n}{n}-\sum_{n=1}^{\infty}\frac{a_n}{n}x^n\right) dx \\ = 2\int_0^1 \frac{\ln(\sqrt{1-x}+1)}{1-x} \ dx=2\int_0^1 \frac{\ln(\sqrt{x}+1)}{x} \ dx ,\end{aligned}

by Problem 1, i). So the substitution $x=t^2$ gives

\displaystyle \begin{aligned} \sum_{n=1}^{\infty}\frac{H_na_n}{n}=4 \int_0^1\frac{\ln(t+1)}{t} \ =4\int_0^1 \sum_{n=1}^{\infty}(-1)^{n-1} \frac{t^{n-1}}{n} \ dt=4\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{3},\end{aligned}

by Problem 1, i), in this post. $\Box$

Exercise. Show that the interval of convergence of $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n}x^n$ is $[-1,1].$

# Harmonic numbers (2)

See the first part of this post here.

Problem. Let $\text{Li}_{\alpha}$ and $\zeta$ be the polylogarithm and zeta functions, respectively. Show that

i) if $\alpha \ge 2,$ then $\displaystyle \lim_{x\to1^-}(\zeta(\alpha)-\text{Li}_{\alpha}(x))\ln(1-x)=0$

ii) $\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n^2}=2\zeta(3)$

iii) $\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n^3}=\frac{\pi^4}{72}.$

Solution. i) For $\alpha > 2,$ we have

\displaystyle \begin{aligned} \lim_{x\to1^-}(\zeta(\alpha)-\text{Li}_{\alpha}(x))\ln(1-x)=\lim_{x\to1^-} \frac{\zeta(\alpha)-\text{Li}_{\alpha}(x)}{1-x} \lim_{x\to1^-} (1-x)\ln(1-x) \\ =\text{Li}_{\alpha}'(1) \lim_{x\to0^+}x \ln x= 0,\end{aligned}

because $\displaystyle \lim_{x\to0^+}x \ln x=0$ and $\displaystyle \text{Li}_{\alpha}'(1)=\text{Li}_{\alpha -1}(1)= \zeta(\alpha -1)$ is a finite number.
For $\alpha=2,$ we use Problem 1 in this post to get

\displaystyle \begin{aligned} \lim_{x\to1^-}(\zeta(2)-\text{Li}_2(x))\ln(1-x)= \lim_{x\to1^-}\left(\frac{\pi^2}{6}-\text{Li}_2(x)\right)\ln(1-x) \\ =\lim_{x\to1^-}\left(\text{Li}_2(1-x)+\ln x \ln(1-x)\right)\ln(1-x) \\ =\lim_{x\to0^+}\left(\text{Li}_2(x)+\ln x \ln(1-x)\right)\ln x=0,\end{aligned}

because $\displaystyle \lim_{x\to0^+} \frac{\text{Li}_2(x)}{x}=1$ and so $\displaystyle \lim_{x\to0^+}\text{Li}_2(x) \ln x =\lim_{x\to0^+}\frac{\text{Li}_2(x)}{x} \lim_{x\to0^+} x \ln x =0$ and we also have $\displaystyle \lim_{x\to0^+}(\ln x)^2\ln(1-x)=0.$

ii) We have $\displaystyle H_n=\int_0^1 \frac{1-x^n}{1-x} \ dx,$ by the remark in part (1), and thus

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{H_n}{n^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}\int_0^1 \frac{1-x^n}{1-x} \ dx=\int_0^1 \frac{1}{1-x}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{x^n}{n^2}\right) dx \\ =\int_0^1 \frac{1}{1-x}(\zeta(2)-\text{Li}_2(x)) dx,\end{aligned}

Now we are going to use integration by parts with $\displaystyle \zeta(2)-\text{Li}_2(x)=u$ and $\displaystyle \frac{dx}{1-x}=dv.$ Since

$\displaystyle \text{Li}_2'(x)=\frac{\text{Li}_1(x)}{x}=-\frac{\ln(1-x)}{x},$

we have $\displaystyle du = \frac{\ln(1-x)}{x} \ dx$ and $v=-\ln(1-x).$ Thus, since, by i), $\displaystyle \lim_{x\to1^{-}}uv=0,$ we have

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{H_n}{n^2}=\int_0^1 \frac{(\ln(1-x))^2}{x} \ dx=\int_0^1 \frac{(\ln x)^2}{1-x} \ dx=\int_0^{\infty} \frac{t^2e^{-t}}{1-e^{-t}} \ dt =2\zeta(3), \ \ \ \ x=e^{-t},\end{aligned}

by Problem 2, i), in this post.

iii) The idea for this part is the same as ii). We have

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{H_n}{n^3}=\sum_{n=1}^{\infty}\frac{1}{n^3}\int_0^1 \frac{1-x^n}{1-x} \ dx=\int_0^1 \frac{1}{1-x}\left(\sum_{n=1}^{\infty}\frac{1}{n^3}-\sum_{n=1}^{\infty}\frac{x^n}{n^3}\right) dx \\ =\int_0^1 \frac{1}{1-x}(\zeta(3)-\text{Li}_3(x)) dx.\end{aligned}

We now use integration by parts with $\displaystyle \zeta(3)-\text{Li}_3(x)=u$ and $\displaystyle \frac{dx}{1-x}=dv.$ Since $\displaystyle \text{Li}_3'(x)=\frac{\text{Li}_2(x)}{x},$ we have $\displaystyle du = -\frac{\text{Li}_2(x)}{x} \ dx$ and $v=-\ln(1-x)$ and thus, since, by i), $\displaystyle \lim_{x\to1^{-}}uv=0,$ we have

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{H_n}{n^3}=-\int_0^1 \frac{\ln(1-x)\text{Li}_2(x)}{x} \ dx=\int_0^1\text{Li}_2'(x)\text{Li}_2(x) \ dx=\frac{1}{2}(\text{Li}_2(x))^2 \Big\vert_0^1 \\ =\frac{1}{2}(\text{Li}_2(1))^2=\frac{1}{2}(\zeta(2))^2=\frac{\pi^4}{72}. \ \Box \end{aligned}

Exercise 1. Think about this: does the result given in the first part of the above problem hold for $1 < \alpha < 2 ?$

Exercise 2. Show that $\displaystyle \lim_{x\to0^+}x\ln x = \lim_{x\to0^+} (\ln x)^2\ln(1-x)=0.$ These limits were used in the solution of the first part of the above problem.

# Yet another interesting AMM problem

An unnecessarily complicated solution of the following problem is given here. My solution although uses the same idea but keeps it simple.

Problem 1. Show that

$\displaystyle \tan^{-1}x \ln(1+x^2)=2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{2n+1}x^{2n+1}, \ \ \ \ \ x \in [-1,1],$

where $H_n$ is the $n$-th Harmonic number.

Solution. We have

\displaystyle \begin{aligned} \tan^{-1}x \ln(1+x^2)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^{2n+2}=x^3 \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n}\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^{2n} \\ =x^3\sum_{n=0}^{\infty}(-1)^n\sum_{k=0}^n\frac{1}{(k+1)(2n-2k+1)}x^{2n}=2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+3}\sum_{k=0}^n\left(\frac{1}{2k+2}+\frac{1}{2n-2k+1}\right)x^{2n+3} \\ =2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+3}\sum_{k=1}^{2n+2}\frac{1}{k} \ x^{2n+3}=2\sum_{n=0}^{\infty}(-1)^n \frac{H_{2n+2}}{2n+3} \ x^{2n+3}=2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{2n+1} \ x^{2n+1}.\end{aligned}

It is clear that the identity

$\displaystyle \tan^{-1}x \ln(1+x^2)=2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{2n+1}x^{2n+1}$

holds for $x \in (-1,1)$ because the intervals of convergence of the Maclaurin series of $\tan^{-1}x$ and $\ln(1+x^2)$ are both $[-1,1].$ Also, by the alternating test, the identity holds for $x=\pm 1$ (see Exercise 2 in this post!). $\Box$

Problem 2 (American Mathematical Monthly, 2018). Show that

$\displaystyle I:=\int_0^1 \frac{\tan^{-1}x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}.$

Solution. We have

$\displaystyle I=\int_0^1 \frac{\tan^{-1}x}{x}\ln(1+x^2) \ dx - 2\int_0^1 \frac{\tan^{-1}x}{x}\ln(1-x) \ dx.$

Let $\displaystyle J:=\int_0^1 \frac{\tan^{-1}x}{x}\ln(1+x^2) \ dx$ and $\displaystyle K:=\int_0^1 \frac{\tan^{-1}x}{x}\ln(1-x) \ dx.$ So

$I=J-2K. \ \ \ \ \ \ \ \ \ \ \ \ (1)$

By Problem 1, we have

\displaystyle \begin{aligned}J=\int_0^1 2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{2n+1}x^{2n} dx=2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{2n+1}\int_0^1x^{2n}dx \\ =2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{(2n+1)^2}.\end{aligned} \ \ \ \ \ \ \ \ \ (2)

Now, to find $K,$ we do not use the method we used to find $J$ because, unlike the Maclaurin series of $\tan^{-1}x \ln(1+x^2),$ the Maclaurin series of $\tan^{-1}x \ln(1-x)$ is not very nice. Instead, we use the Maclaurin series of $\tan^{-1}x$ and then we find each term, which is in the form $\displaystyle \int_0^1 x^m\ln(1-x) \ dx,$ using  integration by parts. So we have

\displaystyle \begin{aligned}K=\int_0^1\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n}\ln(1-x) \ dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln(1-x) \ dx.\end{aligned} \ \ \ \ \ \ \ \ (3)

We should obviously choose $\ln(1-x) = u$ and $x^{2n}dx = dv$ but we need to be careful here because $\displaystyle \lim_{x\to1^{-}}u=-\infty,$ So we need to choose $v$ such that $\displaystyle \lim_{x\to1^{-}}v=0,$ i.e. we need to choose $\displaystyle v=\frac{x^{2n+1}}{2n+1}-\frac{1}{2n+1}.$ Now see that $\displaystyle \lim_{x\to1^{-}}uv=0$ and hence integration by parts gives

\displaystyle \begin{aligned} \int_0^1x^{2n}\ln(1-x) \ dx=\int_0^1\left(\frac{x^{2n+1}}{2n+1}-\frac{1}{2n+1}\right)\frac{dx}{1-x}=-\frac{1}{2n+1}\int_0^1\frac{1-x^{2n+1}}{1-x} \ dx \\ =-\frac{1}{2n+1}\int_0^1(1+x+ \cdots + x^{2n}) \ dx=-\frac{1}{2n+1}\sum_{k=1}^{2n+1}\frac{1}{k}= -\frac{H_{2n+1}}{2n+1}.\end{aligned}

Thus, by $(3),$

$\displaystyle K=-\sum_{n=0}^{\infty}(-1)^n\frac{H_{2n+1}}{(2n+1)^2}$

and hence, by $(1)$ and $(2),$

\displaystyle \begin{aligned}I=J-2K=2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{(2n+1)^2}+2\sum_{n=0}^{\infty}(-1)^n\frac{H_{2n+1}}{(2n+1)^2} \\ =2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{(2n+1)^2}+2+2\sum_{n=1}^{\infty}(-1)^n\frac{H_{2n+1}}{(2n+1)^2} \\ =2+2\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n+1)^2}(H_{2n+1}-H_{2n}) =2+2\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n+1)^3} \\ =2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{16},\end{aligned}

by Problem 1 in this post. $\Box$

Exercise 1. In Exercise 4 in this post, we saw that $\displaystyle \int_0^{\pi/2}\frac{\ln(1-\sin x)}{\sin x} \ dx = -\frac{3\pi^2}{8}.$ Now show that $\displaystyle \int_0^{\pi/2} \frac{x\ln(1-\sin x)}{\sin x} \ dx = -\frac{\pi^3}{8}.$

Exercise 2. Show that $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n+1}=\frac{1}{4}(\ln 2)^2-\ln 2 + \frac{\pi}{2}-\frac{\pi^2}{16}.$
Hint. Start with integrating the Maclaurin series of $\tan^{-1}x \ln(1+x^2)$ given in Problem 1.

# Harmonic numbers (1)

Definition. Given an integer $n \ge 1,$ the $n$-th harmonic number $H_n$ is defined by

$\displaystyle H_n=\sum_{k=1}^n \frac{1}{k}.$

Remark. i) $\displaystyle H_n=\sum_{k=1}^{n-1}\frac{1}{k}+\frac{1}{n}=H_{n-1}+\frac{1}{n}.$

ii) $\displaystyle \sum_{k=1}^n \frac{1}{2k-1}=\sum_{k=1}^{2n} \frac{1}{k}-\sum_{k=1}^n \frac{1}{2k}=H_{2n}-\frac{1}{2}H_n.$

iii) $\displaystyle \lim_{n\to\infty}(H_n-\ln n)=\gamma,$ where $\gamma$ is the Euler’s constant (see Problem 2 in this post!).

iv) $\displaystyle H_n=\sum_{k=1}^n \frac{1}{k}=\sum_{k=1}^n \int_0^1 x^{k-1} dx=\int_0^1 \sum_{k=1}^n x^{k-1} dx = \int_0^1 \frac{1-x^n}{1-x} \ dx.$

v) $\displaystyle \ln(n+1) < H_n < 1+\ln n$ for $n \ge 2.$ This easily follows from estimating the area under the curve $\displaystyle y=\frac{1}{x}$ using rectangles.

vi) $\displaystyle \sum_{n=1}^{\infty}H_nx^n=\frac{\ln(1-x)}{x-1}$ for $x \in (-1,1).$ This is easy to see because

$\displaystyle \frac{-\ln(1-x)}{1-x}=\sum_{n=1}^{\infty}\frac{x^n}{n} \sum_{n=0}^{\infty}x^n=\sum_{n=1}^{\infty}H_n x^n.$

Problem 1. (Maclaurin series of $(\ln(1-x))^2$ and $(\tan^{-1}x)^2$). Show that

i) $\displaystyle (\ln(1-x))^2=2\sum_{n=2}^{\infty}\frac{H_{n-1}}{n} x^n$ for $x \in [-1,1)$

ii) $\displaystyle (\tan^{-1}x)^2=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(H_{2n}-\frac{1}{2}H_n \right)x^{2n}$ for $x \in [-1,1].$

Solution. i) By part vi) of the above remark, $\displaystyle \sum_{n=1}^{\infty} H_nx^n=\frac{\ln(1-x)}{x-1},$ for $x \in (-1,1),$ and so integrating gives

$\displaystyle \sum_{n=2}^{\infty} \frac{H_{n-1}}{n} x^n=\sum_{n=1}^{\infty} \frac{H_n}{n+1}x^{n+1}=\frac{1}{2}(\ln(1-x))^2$

for $x \in (-1,1)$ and the result follows. The only thing left is to show that $\displaystyle \sum_{n=2}^{\infty}\frac{H_{n-1}}{n} x^n$ is convergent at $x=-1$ and that follows from the alternating test (see Exercise 2 for a hint!).

ii) For $x \in (-1,1)$ we have

\displaystyle \begin{aligned} (\tan^{-1}x)^2=2\int_0^x \frac{\tan^{-1}t}{1+t^2} \ dt=2\int_0^x \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}t^{2n+1} \sum_{n=0}^{\infty}(-1)^nt^{2n} \\ = 2\int_0^x \sum_{n=1}^{\infty}(-1)^{n-1} \sum_{k=1}^n \frac{1}{2k-1}t^{2n-1} dt=2\sum_{n=1}^{\infty} (-1)^{n-1}\sum_{k=1}^n \frac{1}{2k-1} \int_0^x t^{2n-1}dt \\ =2\sum_{n=1}^{\infty} (-1)^{n-1}\left(H_{2n}-\frac{1}{2}H_n\right)\frac{x^{2n}}{2n}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(H_{2n}-\frac{1}{2}H_n\right)x^{2n}. \end{aligned}

The only thing left to prove is that $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(H_{2n}-\frac{1}{2}H_n \right)x^{2n}$ is convergent at $x=\pm 1$ and that follows from the alternating test (see Exercise 3). $\Box$

Problem 2. The polylogarithm function $\displaystyle \text{Li}_{\alpha}(x):=\sum_{n=1}^{\infty} \frac{x^n}{n^{\alpha}}$ was introduced in this post. Show that

i) $\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n}x^n=\frac{1}{2}(\ln(1-x))^2+\text{Li}_2(x),$ for $x \in [-1,1)$

ii) $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}x^n=(\tan^{-1}\sqrt{x})^2-\frac{1}{4}(\ln(1+x))^2-\frac{1}{2}\text{Li}_2(-x),$ for $x \in [0,1].$

Solution. i) By Problem 1, i) and the first part of the above remark, we have

\displaystyle \begin{aligned}(\ln(1-x))^2=2\sum_{n=2}^{\infty} \frac{1}{n}\left(H_n - \frac{1}{n}\right)x^n=2\sum_{n=2}^{\infty} \frac{H_n}{n}x^n-2(\text{Li}_2(x)-x) \\ =2\sum_{n=1}^{\infty} \frac{H_n}{n}x^n-2\text{Li}_2(x) \end{aligned}

and the result follows.

ii) By Problem 1, ii), we have

$\displaystyle (\tan^{-1}\sqrt{x})^2=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}x^n+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_n}{n}(-x)^n$

and the result follows from i). $\Box$

Example. Show that

i) $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n}{n}=\frac{\pi^2}{12}-\frac{1}{2}(\ln 2)^2.$

ii) $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}=\frac{5\pi^2}{48}-\frac{1}{4}(\ln 2)^2.$

iii) $\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n 2^n}=\frac{\pi^2}{12}.$

Solution. For i), put $x=-1$ in Problem 2, i), and for ii), put $x=1$ in Problem 2, ii). Notice that by Problem 1, i), in this post,

$\displaystyle \text{Li}_2(-1)=\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}=-\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}=\left(1-\frac{1}{2}\right)\zeta(2)=-\frac{\pi^2}{12}.$

To prove iii), put $x=1/2$ in Problem 2, ii), to get

$\displaystyle \sum_{n=1}^{\infty}\frac{H_n}{n2^n}=\frac{1}{2}(\ln 2)^2+\text{Li}_2(1/2)=\frac{\pi^2}{12},$

by Problem 1 in this post. $\Box$

Exercise 1. Given positive integers $a,b,$ show that $\displaystyle \lim_{n\to\infty}(H_{na}-H_{nb})=\ln(a/b).$
Hint. $H_{na}-H_{nb}=H_{na}-\ln (na) - (H_{nb}-\ln (nb))+\ln(a/b).$

Exercise 2. Consider the sequence $\displaystyle a_n:=\frac{H_n}{n+1}.$ Show that $\{a_n\}$ is decreasing and $\displaystyle \lim_{n\to\infty}a_n=0.$
Hint. It’s easy to directly show that $a_n \ge a_{n+1}$ and $\displaystyle \lim_{n\to\infty}a_n=0$ by the part iii) or vi) of the above remark.

Exercise 3. Consider the sequence $\displaystyle b_n:=\frac{1}{n}\sum_{k=1}^n \frac{1}{2k-1}=\frac{1}{n}\left(H_{2n}-\frac{1}{2}H_n\right).$ Show that $\{b_n\}$ is decreasing and $\displaystyle \lim_{n\to\infty}b_n=0.$

# Convergence of series (4)

Problem. Let $f:[0,1] \longrightarrow \mathbb{R}$ be a function such that $f,f',f''$ are all continuous. Show that the series $\displaystyle \sum_{n=1}^{\infty} f(1/n)$ is convergent if and only if $f(0)=f'(0)=0.$

Solution. By Taylor’s theorem, for any integer $n \ge 1,$ there exists $c_n \in (0,1/n)$ such that

$\displaystyle f(1/n)=f(0)+\frac{f'(0)}{n}+\frac{f''(c_n)}{2n^2}. \ \ \ \ \ \ \ \ \ \ (1)$

Also, since since $f''$ is continuous, there exists a real constant $M$ such that

$|f''(x)| \le M, \ \ \ \forall x \in [0,1]. \ \ \ \ \ \ \ \ \ \ (2)$

Now suppose that $\displaystyle \sum_{n=1}^{\infty} f(1/n)$ is convergent. Then since $f$ is continuous, we must have

$\displaystyle 0=\lim _{n\to\infty} f(1/n)=f(\lim_{n\to\infty} 1/n)=f(0).$

So $(1)$ becomes

$\displaystyle f(1/n)=\frac{f'(0)}{n}+\frac{f''(c_n)}{2n^2}. \ \ \ \ \ \ \ \ \ \ \ (3)$

But $\displaystyle \sum_{n=1}^{\infty} \frac{|f''(c_n)|}{2n^2} \le \frac{M}{2}\sum_{n=1}^{\infty} \frac{1}{n^2},$ by $(2),$ and so, by the comparison test, $\displaystyle \sum_{n=1}^{\infty} \frac{f''(c_n)}{2n^2}$ is convergent. Hence the series $\displaystyle f'(0)\sum_{n=1}^{\infty} \frac{1}{n}$ must converge too because, by $(3),$

$\displaystyle f'(0)\sum_{n=1}^{\infty} \frac{1}{n}=\sum_{n=1}^{\infty}f(1/n) - \sum_{n=1}^{\infty} \frac{f''(c_n)}{2n^2}.$

But since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges, $\displaystyle f'(0)\sum_{n=1}^{\infty} \frac{1}{n}$ converges only if $f'(0)=0.$

Conversely, suppose that $f(0)=f'(0)=0.$ Then $(1)$ becomes

$\displaystyle f(1/n) = \frac{f''(c_n)}{2n^2}$

and hence $\displaystyle |f(1/n)| \le \frac{M}{2n^2},$ by $(2).$ Thus the series $\displaystyle \sum_{n=1}^{\infty} f(1/n)$ is convergent, by the comparison test. $\Box$

# Integral of sin(x)/x – an example

Here I gave an elementary proof of $\displaystyle \int_0^{\infty} \frac{\sin x}{x} \ dx = \frac{\pi}{2}.$ Thus, changing the variable $x$ to $ax,$ where $a > 0$ is any real constant, gives $\displaystyle \int_0^{\infty} \frac{\sin(ax)}{x} \ dx = \frac{\pi}{2}.$
The following example is some kind of extension of the above result.

Example. Let $a_0, a_1, \cdots , a_n, \ n \ge 1,$ be positive real numbers such that $\displaystyle a_0 > \sum_{k=1}^na_k.$ Show that $\displaystyle \int_0^{\infty} \frac{\sin(a_0x)\sin(a_1x) \cdots \sin(a_nx)}{x^{n+1}} \ dx = \frac{\pi}{2}a_1a_2 \cdots a_n.$

Solution. The solution is by induction over $n.$ Suppose first that $n=1.$ We use integration by parts with $\sin(a_0x)\sin(a_1x)=u$ and $\displaystyle \frac{dx}{x^2}=dv$ to get

\displaystyle \begin{aligned} \int_0^{\infty} \frac{\sin(a_0x)\sin(a_1x)}{x^2} \ dx = a_0\int_0^{\infty} \frac{\cos(a_0x)\sin(a_1x)}{x} \ dx + a_1\int_0^{\infty} \frac{\cos(a_1x)\sin(a_0x)}{x} \ dx \\ \\ =\frac{a_0}{2}\int_0^{\infty} \frac{\sin((a_0+a_1)x)-\sin((a_0-a_1)x)}{x} \ dx + \frac{a_1}{2}\int_0^{\infty} \frac{\sin((a_0+a_1)x)+\sin((a_0-a_1)x)}{x} \ dx \\ \\ = \frac{a_0}{2}\left(\frac{\pi}{2}-\frac{\pi}{2}\right)+\frac{a_1}{2}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)=\frac{\pi}{2}a_1. \end{aligned}

Now suppose that $\displaystyle n \ge 2$ and the claim is true for $n-1.$ Then integration by parts with $\displaystyle \prod_{k=0}^n \sin(a_kx)=u$ and $\displaystyle \frac{dx}{x^{n+1}}=dv$ gives

\displaystyle \begin{aligned} \int_0^{\infty}\prod_{k=0}^n \frac{\sin(a_kx)}{x} \ dx = \frac{1}{n}\int_0^{\infty}\frac{1}{x^n}\sum_{k=0}^n a_k\cos(a_kx)\prod_{0 \le j \le n, \ j \ne k}\sin(a_jx) \ dx \\ =\frac{a_0}{2n}\int_0^{\infty}\frac{1}{x^n}(\sin((a_0+a_1)x)-\sin((a_0-a_1)x))\prod_{j=2}^n\sin(a_jx) \ dx+ \\ \sum_{k=1}^{\infty} \frac{a_k}{2n}\int_0^{\infty}\frac{1}{x^n} (\sin((a_0+a_k)x) + \sin((a_0-a_k)x)\prod_{1 \le j \le n, \ j \ne k} \sin(a_jx) \ dx.\end{aligned}

But, by our induction hypothesis, for $1 \le k \le n$ we have

$\displaystyle \int_0^{\infty}\frac{1}{x^n}\sin((a_0 \pm a_k)x)\prod_{1 \le j \le n, \ j \ne k}\sin(a_jx) \ dx = \frac{\pi}{2}\prod_{j \neq k}a_j$

because $a_k > 0$ and $\displaystyle a_0 > \sum_{k=1}^n a_k$ implies that $\displaystyle a_0 \pm a_k > \sum_{1 \le j \le n, \ j \ne k}a_j.$
Thus

\displaystyle \begin{aligned} \int_0^{\infty}\prod_{k=0}^n \frac{\sin(a_kx)}{x} \ dx=\sum_{k=1}^{\infty} \frac{a_k}{2n}\int_0^{\infty}\frac{1}{x^n} (\sin((a_0+a_k)x) + \sin((a_0-a_k)x)\prod_{1 \le j \le n, \ j \ne k} \sin(a_jx) \ dx \\ = \sum_{k=1}^n \frac{\pi a_k}{2n}\prod_{1 \le j \le n, \ j \ne k}a_j=\frac{\pi}{2}\prod_{j=1}^na_j\end{aligned}

and that completes the induction. $\Box$

# Convergence of series (3)

Problem. Let $\{a_n\}_{n \ge 1}$ be a sequence of positive real numbers and let $\displaystyle b_n:=\frac{\ln n}{1+\ln n}, \ n \ge 1.$ Show that if $\displaystyle \sum_{n=1}^{\infty}a_n$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_n^{b_n}$ converges too.

Solution. Let $\displaystyle u_n:=a_n^{b_n}$ and consider the set

$\displaystyle I:=\{n \ge 1: \ \ u_n > e^2a_n\}.$

If $n \in I,$ then

$\displaystyle u_n^{\ln n} > (e^2a_n)^{\ln n}=e^{2\ln n}a_n^{\ln n}=n^2a_n^{(1+\ln n)b_n}=n^2u_n^{1+\ln n}$

and hence $\displaystyle u_n < \frac{1}{n^2}.$ Thus

$\displaystyle 0 < \sum_{n=1}^{\infty} u_n=\sum_{n \in I}u_n + \sum_{n \notin I} u_n < \sum_{n=1}^{\infty} \frac{1}{n^2}+e^2\sum_{n=1}^{\infty}a_n$

and the result follows from the comparison test because both $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ and $\displaystyle \sum_{n=1}^{\infty}a_n$ are convergent. $\Box$

# Integral of sin(x)/x

There are several proofs of the well-known result $\displaystyle \int_0^{\infty} \frac{\sin x}{x} \ dx=\frac{\pi}{2}$ but most of them are beyond the first course in calculus. In this post, we are going to see a very elementary proof of that result. The proof is a fairly straightforward result of the following problem.

Problem 1. Show that $\displaystyle \lim_{a\to{+\infty}}\frac{1}{a}\int_0^{\pi/2}\left(\frac{1}{x^2}-\frac{\cos x}{\sin^2x}\right)\cos(ax) \ dx = 0.$

Solution. See that $\displaystyle \lim_{x\to{0^+}}\left(\frac{1}{x^2}-\frac{\cos x}{\sin^2x}\right)=\frac{1}{6}.$ So the function

$\displaystyle f(x):=\begin{cases} \frac{1}{x^2}-\frac{\cos x}{\sin^2x} & x \in (0,\pi/2] \\ \frac{1}{6} & x=0 \end{cases}$

is continuous on $[0,\pi/2]$ and thus, by the extreme value theorem, $|f(x)| \le M$ for some real number $M$ and all $x \in [0,\pi/2].$ Thus

$-M \le f(x)\cos(ax) \le M$

for all $x \in [0,\pi/2]$ and hence

$\displaystyle -\frac{M\pi}{2a} \le \frac{1}{a}\int_0^{\pi/2}f(x)\cos(ax) \ dx \le \frac{M\pi}{2a}$

for $a > 0.$ So, by the squeeze theorem, $\displaystyle \lim_{a\to{+\infty}}\frac{1}{a}\int_0^{\pi/2}f(x) \cos(ax) \ dx = 0. \ \Box$

Problem 2. Show that $\displaystyle \int_0^{\infty} \frac{\sin x}{x} \ dx=\frac{\pi}{2}.$

Solution. Let $n \ge 1$ be an integer. Here (Problem 1) we showed that

$\displaystyle \sum_{k=0}^n \cos(2kx)= \frac{1}{2} + \frac{\sin((2n+1)x)}{2 \sin x},$

which gives

$\displaystyle \sum_{k=1}^n \cos(2kx)= -\frac{1}{2} + \frac{\sin((2n+1)x)}{2 \sin x}. \ \ \ \ \ \ \ \ \ (1)$

Let $\displaystyle I_n:=\int_0^{\pi/2}\frac{\sin((2n+1)x)}{\sin x} \ dx.$ Then integrating $(1)$ over the interval $[0,\pi/2]$ gives

$\displaystyle 0=\sum_{k=1}^n \int_0^{\pi/2}\cos(2kx) \ dx=-\frac{\pi}{4} + \frac{1}{2}I_n$

and therefore

$\displaystyle I_n=\frac{\pi}{2}. \ \ \ \ \ \ \ \ \ \ (2)$

Let $\displaystyle J_n:=\int_0^{\pi/2}\frac{\sin((2n+1)x)}{x} \ dx.$ Then

$\displaystyle I_n-J_n=\int_0^{\pi/2} \left(\frac{1}{\sin x}-\frac{1}{x}\right)\sin((2n+1)x) \ dx.$

So integration by parts with $\displaystyle \frac{1}{\sin x}-\frac{1}{x}=u$ and $\sin((2n+1)x) \ dx = dv$ gives

$\displaystyle I_n-J_n=\frac{1}{2n+1}\int_0^{\pi/2}\left(\frac{1}{x^2}-\frac{\cos x}{\sin^2x}\right)\cos((2n+1)x) \ dx.$

Thus, by Problem 1 and $(2),$ we have

\displaystyle \begin{aligned} \frac{\pi}{2}-\lim_{n\to\infty}J_n=\lim_{n\to\infty}(I_n-J_n)=\lim_{n\to\infty}\frac{1}{2n+1}\int_0^{\pi/2}\left(\frac{1}{x^2}-\frac{\cos x}{\sin^2x}\right)\cos((2n+1)x) \ dx \\ =0\end{aligned}

and hence

$\displaystyle \lim_{n\to\infty}J_n=\frac{\pi}{2}. \ \ \ \ \ \ \ \ \ \ (3)$

Finally, the substitution $(2n+1)x=t$ gives

$\displaystyle J_n=\int_0^{\pi/2}\frac{\sin((2n+1)x)}{x} \ dx= \int_0^{n\pi+\pi/2}\frac{\sin t}{t} \ dt$

and thus, since $\displaystyle \int_0^{\infty} \frac{\sin t}{t} \ dt$ is convergent (see this post!), we have

$\displaystyle \int_0^{\infty} \frac{\sin t}{t} \ dt = \lim_{n\to\infty} \int_0^{n\pi + \pi/2} \frac{\sin t}{t} \ dt = \lim_{n\to\infty}J_n=\frac{\pi}{2},$

by $(3). \ \Box$

# Divergence of the sequence of sine of Fibonacci numbers

Throughout this post, $\{F_n\}$ is the Fibonacci sequence. I’m going to prove that the sequence $\{\sin(F_n)\}$ is divergent, i.e. $\displaystyle \lim_{n\to\infty} \sin(F_n)$ does not exist.

Notation. Let $\mathbb{Q}$ be the set of rational numbers. We denote by $\mathbb{Q}[\sqrt{5}]$ the set $\{p+q\sqrt{5}: \ \ p,q \in \mathbb{Q}\}.$ For example, $\displaystyle \left(\frac{\sqrt{5}+1}{2}\right)^n \in \mathbb{Q}[\sqrt{5}]$ for all integers $n.$

Remark 1. See that $\mathbb{Q} \subset \mathbb{Q}[\sqrt{5}]$ and if $a,b \in \mathbb{Q}[\sqrt{5}],$ then $a \pm b, ab \in \mathbb{Q}[\sqrt{5}].$ Also if $a,b \in \mathbb{Q}[\sqrt{5}]$ and $b \ne 0,$ then $\displaystyle \frac{a}{b} \in \mathbb{Q}[\sqrt{5}].$

Remark 2. It is clear that every element of $\displaystyle \mathbb{Q}[\sqrt{5}]$ is a root of some quadratic polynomial with integer coefficients. We showed here that $\pi \notin \mathbb{Q}$ and we also mentioned، without a proof,  that in fact $\pi$ is not a root of any polynomial with integer  coefficients (see the Remark in that post!). In particular, $\pi \notin \mathbb{Q}[\sqrt{5}].$

Problem. Let $\displaystyle L:=\lim_{n\to\infty} \sin(F_n).$ Show that

i) if $L$ exists, then $L=0$

ii) $L$ does not exist.

Solution. i) Suppose that $L \ne 0.$ Since $F_{n+2}=F_{n+1}+F_n$ and $F_{n-1}=F_{n+1}-F_n,$ we have

$\displaystyle \sin F_{n+2}-\sin F_{n-1}=2\sin F_n \cos F_{n+1}, \ \ \ \ \ \ \ \ (1)$

$\displaystyle \sin F_{n+2}+\sin F_{n-1}=2\sin F_{n+1}\cos F_n. \ \ \ \ \ \ \ \ \ (2)$

Now, since $\displaystyle \lim_{n\to\infty}(\sin F_{n+2}-\sin F_{n-1})=L-L=0,$ we must have $\displaystyle \lim_{n\to\infty}\cos F_n=0,$ by $(1),$ and so, by $\displaystyle (2), \ 2L=\lim_{n\to\infty}(\sin F_{n+2}+\sin F_{n-1})=0,$ contradiction.

ii) Suppose that $L$ exists. Then by i), $L=0.$ For each $n,$ there exists positive integer $k_n$ and $\theta_n \in (-\pi/2, \pi/2)$ such that

$F_n=k_n \pi + \theta_n.$

Since $\displaystyle \lim_{n\to\infty}\sin F_n=0,$ we must have $\displaystyle \lim_{n\to\infty}\sin \theta_n=0$ and hence $\displaystyle \lim_{n\to\infty}\theta_n=0.$ Thus

$\displaystyle \lim_{n\to\infty}(k_{n+1}-k_n-k_{n-1})=\frac{1}{\pi}\lim_{n\to\infty}(\theta_{n-1}+\theta_n-\theta_{n+1})=0.$

So, since $k_n$ are all integers, there exists an integer $m$ such that

$k_{n+1}=k_n+k_{n-1}$

for $n \ge m.$ Thus, by Problem 1 in this post, there exist constants $a,b \in \mathbb{Q}[\sqrt{5}]$ such that

$k_{n+m}=a\varphi^n+b \psi^n, \ \ \ \forall n \ge 0,$

where $\displaystyle \varphi=\frac{\sqrt{5}+1}{2}$ and $\displaystyle \psi=\frac{1-\sqrt{5}}{2}=-\frac{1}{\varphi}.$ Therefore

$F_{n+m}=k_{n+m}\pi + \theta_{n+m}=\pi(a\varphi^n+b\psi^n)+\theta_{n+m}$

and hence

$\displaystyle \frac{F_{n+m}}{\varphi^n}=\pi a + \pi b\frac{\psi^n}{\varphi^n}+\frac{\theta_{n+m}}{\varphi^n},$

which, after taking limit as $n\to\infty,$ gives $\displaystyle \frac{\varphi^m}{\sqrt{5}}=\pi a.$ So $\displaystyle \pi=\frac{\varphi^m}{a\sqrt{5}} \in \mathbb{Q}[\sqrt{5}],$ which is false, by Remark 2. So $L \ne 0$ and thus $L$ doesn’t exist. $\Box$

Exercise. Show that $\displaystyle \lim_{n\to\infty} \cos(F_n)$ does not exist.