# Convergence of $\sum_{n=2}^\infty \frac{1}{nn^{1/n}}$

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 $$L\rightarrow \infty$$?

• Compare with $\sum 1/n$ and conclude. – Paramanand Singh May 26 at 5:49
• Hint: $\frac{1}{{n^{1/n} }} > \frac{1}{2}.$ – Gary May 26 at 5:55

There are plenty of proofs that $$2^n>n,$$ for all integers $$n\geq 0.$$ Thus $$0 and thus:

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

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.

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.