# Test for convergence the series $\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}$

Test for convergence the series $$\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}$$ I'd like to make up a collection with solutions for this series, and any new
solution will be rewarded with upvotes. Here is what I have at the moment

Method 1

We know that for all positive integers $n$, $n<2^n$, and this yields $$n^{(1/n)}<2$$ $$n^{(1+1/n)}<2n$$ Then, it turns out that $$\frac{1}{2} \sum_{n=1}^{\infty}\frac{1}{n} \rightarrow \infty \le\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}$$ Hence $$\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}\rightarrow \infty$$ EDIT:

Method 2

If we consider the maximum of $f(x)=x^{(1/x)}$ reached for $x=e$
and denote it by $c$, then $$\sum_{n=1}^{\infty}\frac{1}{c \cdot n} \rightarrow \infty \le\sum_{n=1}^{\infty}\frac{1}{n^{(n+1)/n}}$$

Thanks!

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Let $a_n = 1/n^{(n+1)/n}$. Then $$\begin{eqnarray*} \frac{a_n}{a_{n+1}} &\sim& 1+\frac{1}{n} - \frac{\log n}{n^2} \qquad (n\to\infty). \end{eqnarray*}$$ The series diverges by Bertrand's test.

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beautiful and fast as usual! (+1) Glad to learn things from you. :-) – user 1618033 Dec 28 '12 at 21:57
@Chris'ssister: Glad to help. Yet another interesting question. (+1) – user26872 Dec 28 '12 at 21:58
I think that you should not write this with the $\sim$ symbol (here, it only means that the limit is 1). – Siméon Dec 31 '12 at 16:06
@JulienB.: This is an asymptotic expansion. A standard meaning of the symbol $\sim$ is "asymptotic to." The terms written indeed show the limit is 1 but, more importantly, that the series diverges by Bertand's test. – user26872 Dec 31 '12 at 18:48
@oen: I did not know this meaning. To me, $a_n \sim b_n$ means that $a_n/b_n \to 1$ as $n \to \infty$. – Siméon Jan 1 '13 at 14:49

With an equivalent: $$\frac{1}{n^{(n+1)/n}} = \exp\left(-\left(1+\frac{1}{n}\right)\ln n\right) = e^{-\ln n+o(1)} \sim \frac{1}{n}.$$

With an inequality (using $\ln(1+x) \leq x$): $$\dfrac{n}{n^{(n+1)/n}} = \exp\left(-\frac{\ln n}{n}\right) \geq \exp\left(-\frac{n-1}{n}\right) = \exp\left(\frac{1}{n}-1\right) \geq e^{-1}.$$ Hence, $$\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^{(n+1)/n}} \geq \dfrac{1}{e} \sum_{n=1}^\infty\dfrac{1}{n} = +\infty.$$

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Thank you for your solutions (+1) – user 1618033 Dec 28 '12 at 17:36

Method 1: Limit comparison with $\frac{1}{n}$

Method 2: $$\frac{1}{n^{1+1/n}}\leq\frac{1}{(n+1)^{1+\frac{1}{n+1}}}$$ Now use Cauchy condensation test, to get that syor sum converges iff the following sum converges: $$\sum_{n\in Z^+}2^n\frac{1}{(2^n)^{1+2^{-n}}}=\sum_{n\in Z^+}\frac{1}{2^{n2^{-n}}}$$

Since $n<2^n$, therefore $n2^{-n}<1$. Thus, $\frac{1}{2}<\frac{1}{2^{n2^{-n}}}$

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Thank you for your solutions (+1) – user 1618033 Dec 28 '12 at 17:36