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I am trying to determine the convergence of the series:

$$ \sum_{n=1}^{\infty} \frac{n^{n}}{(n+1)^{n+1}} = 1 + \frac{1^1}{2^2} + \frac{2^2}{3^3} + \frac{3^3}{4^4} + \frac{4^4}{5^5} + \dots \text{(to infinity)} $$

I've attempted to apply D'Alembert's Ratio Test and Raabe's Test to ascertain convergence, but both tests have proven inconclusive for this series.

Can someone suggest an alternative convergence test or method to determine the convergence or divergence of this series?

Thank you in advance for your help!

Math Attack
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Shreyaskar Saha
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  • The terms are larger than $C/n$, for an appropriately chosen positive $C$; so it diverges by comparison with the harmonic series. – mjqxxxx Nov 18 '23 at 16:34

2 Answers2

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Hint

$$\frac{n^n}{(n+1)^{n+1}}=\frac{1}{n+1}\left(\frac{n}{n+1}\right)^n.$$ Moreover, the sequence $\left(\left(\frac{n}{n+1}\right)^n\right)_{n\in\mathbb N}$ is decreasing and converges to $\frac{1}{e}$. I let you conclude.

Surb
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As the general term will look a lot "like" $1\over n$ for $n$ large, it's fair to try the limit comparison test picking $b_{n} = {1\over n}$.

Set $a_n = \frac{n^{n}}{(n+1)^{n+1}}$, we have that:

$$ \lim_{n\to \infty} {a_n\over b_n} = \lim_{n\to \infty} \frac{n^{n}}{(n+1)^{n+1}} n = \lim_{n\to \infty} \frac{n^{n+1}}{(n+1)^{n+1}} = 1 \in (0,\infty) $$

So both $\sum a_n$ and $\sum b_n$ have the same behavior, and as $\sum b_n$ diverges also $\sum a_n$ does.

Daniele
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