# How to prove convergence of $\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^n$?

How to prove convergence of the following?

$$\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^n$$

Is the following statement true? $$\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^n < \infty$$

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Here is an answer to a similar problem that gets a bit more to the core of the answer: math.stackexchange.com/a/597781/3584 – aaaaaaaaaaaa Dec 8 '13 at 9:31

## 4 Answers

It does not converge. Look at $$\lim_{n\to \infty} \left(\frac{n}{n+1}\right)^n=\frac{1}{e}\neq 0$$ So the series cannot converge.

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So, this series is divergent. $$\sum_{n=1}^∞ {\left({\frac{n}{n+1}}\right)}^n = \infty?$$ – user110037 Dec 7 '13 at 21:57
@user110037 Yes, this is by what is called here the 'limit of the summand' test. It is usually the first one you want to do because it is a simple task to do and easily checks to see if something is necessarily divergent. Though if the limit is $0$, it could still be divergent, take the harmonic series! Anyway, your series again is divergent by the first test listed here: en.wikipedia.org/wiki/Convergence_tests – mathematics2x2life Dec 7 '13 at 21:59

Hint: What is the limit of the sequence $\left(\left(\dfrac{n}{n+1}\right)^n\right)_{n\in \Bbb N}$?

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Note that $$\left(\dfrac{n}{n+1}\right)^n = \dfrac1{\left(1+\dfrac1n\right)^n} \sim \dfrac1{e}$$ and conclude what you want.

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@GitGud Ok. Stupid error. Thanks. Corrected. – user17762 Dec 7 '13 at 21:52
All the cowards can now remove their down-votes. – user17762 Dec 7 '13 at 21:54
-1 For calling me a coward. And for worrying more about speed than correctness of answer. – aaaaaaaaaaaa Dec 7 '13 at 22:03
@eBusiness Ok. Thanks for identifying yourself. – user17762 Dec 7 '13 at 22:05

If a series $\sum_{n=1}^\infty a_n$ converges, then $a_n\to 0$. Which is equivalent to saying that if $a_n$ does not tend to zero, then the corresponding series does not converge.

So, it suffices here to observe that $(\frac{n}{n+1})^n\to\frac{1}{e}$.

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