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I'm trying to find out if the series $$\sum_{n=2}^\infty \frac{1}{nn^{1/n}}$$

converges or not. First with the ratio test and then with the integral test.


Ratio test:

$$r=\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=\lim\frac{nn^{1/n}}{(1+n)(1+n)^{1/(n+1)}}$$ $$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$$ $$\lim_{n\rightarrow \infty}n^{1/n}=\lim e^{\ln(n)/n}=1$$ $$\lim_{n\rightarrow \infty}(n+1)^{1/(n+1)}=\lim e^{\ln(n+1)/(n+1)}=e^0=1$$ That means $r=1$ so ratio text is inconclusive.


Integral test $$\int^L \frac{1}{xx^{1/x}}dx=\int^L\frac{1}{xe^{\ln(x)/x}}dx$$ Let $\ln(x)=y$ then $$\int\frac{1}{\exp\left\{ye^{-y}\right\}}dy $$ I don't know, what do now? How to show if this integral will or will not converge as $L\rightarrow \infty$?

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    $\begingroup$ Compare with $\sum 1/n$ and conclude. $\endgroup$ – Paramanand Singh May 26 at 5:49
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    $\begingroup$ Hint: $\frac{1}{{n^{1/n} }} > \frac{1}{2}.$ $\endgroup$ – Gary May 26 at 5:55
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There are plenty of proofs that $2^n>n,$ for all integers $n\geq 0.$ Thus $0<n^{1/n}<2$ and thus:

$$\frac1{nn^{1/n}}>\frac{1}{2n}$$

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Comparison with Harmonic Series

As shown in this answer $$ n^{1/n}\le1+\sqrt{\frac2n}\tag1 $$ For $n=1$, $n^{1/n}=1$. Inequality $(1)$ shows that for $n\ge2$, we have $n^{1/n}\le2$. This means that for $n\ge1$, $n^{1/n}\le2$. Thus, $$ \sum_{n=1}^\infty\frac1{nn^{1/n}}\ge\sum_{n=1}^\infty\frac1{2n}\tag2 $$ which diverges since the Harmonic Series diverges.


Cauchy Condensation Test

Since the terms tend monotonically to $0$, $$ \sum_{n=1}^\infty\frac1{nn^{1/n}}\tag3 $$ converges if and only if $$ \sum_{n=1}^\infty\frac{2^n}{2^n2^{n/2^n}}\tag4 $$ The terms of $(4)$ tend to $1$, so the series diverges by the Term Test.

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As $k$ gets large, the power on $k,$ $1 + \frac1{k},$ approaches $1.$ So, it should make sense to do the limit comparison test with $\sum_{k = 1}^{\infty} \frac1{k}.$

So, consider the limit $\lim_{k \to \infty} \frac{\frac1{k}}{\frac{1}{kk^{1/k}}} = \lim_{k \to \infty} k^{\frac1{k}} = \lim_{k \to \infty} e^{\frac{\ln k}{k}}.$ Because $f(x) = e^x$ is continuous everywhere, we can say this is equal to $e^{\lim_{k \to \infty} \frac{\ln k}{k}} = e^0 = 1.$

Now, because the limit is finite, either both series converge or both series diverge. The harmonic series $\sum_{k = 1}^{\infty} \frac1{k}$ famously diverges, so our series also diverges.

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