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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!

3 Answers 3

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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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    @JulienB.: The higher order ratio tests involve an asymptotic expansion in large $n$. How the ratio approaches 1 tells us whether the series diverges or not. This technique is underappreciated.2013-01-01
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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)2012-12-28
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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)2012-12-28