- #1

SqueeSpleen

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## Homework Statement

The original problem was to find for which real numbers the series

$$

\sum_{n=1}^{\infty} 2^{n} .\left( \dfrac{ n }{ n+1 } \right)^{n} x^{n}

$$

converges.

I used root criterion, it gave me a radius of convergence equal to 1/2. For x=1/2, I showed that the sequence converges to 1/e, so the series cannot be convergent. For x=-1/2 I think it's convergent but were unable to find a concise way to prove it.

## Homework Equations

None really, not that I can think off. Well, I have the following expression that I don't think it can be simplified. This is what I got when I summed the nth and (n+1)th terms.

$$

- \left( \dfrac{n}{n+1} \right)^{n}+ \left( \dfrac{ n+1 }{ n+2 } \right) ^{n+1}

$$

## The Attempt at a Solution

For x=-1/2 the sequence inside the series has the same absolute value than the previous one, so in absolute value the general term approaches 1/e. I think it converges, but I'm not sure how to prove it. The expression that I found trying to group the first term with the second one, the third one with the fourth one and so, is horrible and I don't think I can work with it.

I'm sure that it's easy to prove that if an oscillating series has an associated sequence that converges to a number, and the rate of convergence is good enough, then the series converges. But I'm not sure if you need a good ratio of convergence or it's valid for any convergence.

Is there any example of an oscillating sequence that converges to a real number in absolute value and it's sum is not convergent?

Anyway, I would like to see if there's an elementary way to solve this exercise because the expressions I found were horrible enough to make me think about proving a lemma rather than working with this series in particular.

Thanks for your help.

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