Here I proved that My solution was nice and elementary but not very natural. A more natural solution would begin with the following line

and it would continue as follows: but and so if we can write then the answer would be

Right, but only if we can write it that way. In fact, it turns out that we can!
In general, we have the following result.

Tannery’s theorem. For an integer let be a function of the integer variable Suppose that there exist sequences and where each is independent of such that and for all If is convergent, then

Proof. Let be given. Since is convergent, and so

if is large enough. Note that doesn’t depend on because doesn’t depend on Thus if is large enough and then

Also since doesn’t depend on and so, by if is large enough, then

Therefore, if is large enough and then and give

Finally, since for each and doesn’t depend on we have and so provided that is large enough. Hence, by

for all that are large enough. So by the definition of limit.

Example 1. Using Tannery’s theorem, show that

Solution. We have

Let Then Also, since for real numbers we also have

Clearly does not depend on and is convergent. Thus, by Tannery’s theorem,

Example 2. Let . Using Tannery’s theorem, show that

Solution. By the binomial theorem, Let Then

and thus Also, because

Clearly doesn’t depend on and is convergent (you may, for example, use the ratio test). Thus, by Tannery’s theorem,

Example 3. Show that

Solution. See that, for a fixed we have
We also have (why?) and thus Since is convergent, we can use Tannery’s theorem and write

Exercise (a stronger version of Tannery’s theorem). Show that Tannery’s theorem still holds if, in the theorem, we replace with where is any increasing function of