Convergence of $\sum_{n=1}^\infty \frac{\sqrt[n]{n} - 1}{n}$

Question

I am looking for the convergence of $$\sum_{n=1}^\infty \frac{\sqrt[n]{n} - 1}n$$

I have tried ratio test/ root test and tried to compare with other convergent series but I do not get anywhere. Also, limit comparison of $a_n$ didn't seems to work.

Answer 1

Since $\lim_{x\to 0^+}x\ln x= 0$ we have $\exp(x\ln x) \sim 1+x\ln x$, and hence, $$\frac{n^{1/n} - 1}{n} = \frac1n\left(\exp\left(-\frac1n\ln(\frac1n)\right)-1\right)\sim - \frac1{n^2}\ln(\frac1n) = \frac{\ln n}{n^2}$$ hence the convergence follows by Bertrand criteria or by Cauchy condensation test. or comparison test as follows

$$n^{3/2}\frac{\ln n}{n^2}= \frac{\ln n}{n^{1/2}}\to 0,~~as~~n\to\infty\implies \frac{\ln n}{n^2}\le \frac{C}{n^{3/2}}$$

Answer 2

Use:

$\sqrt[n]{n} \le 1+ \dfrac{2}{\sqrt n}, \in \mathbb {Z^+}, n \ge 1.$

$a_n: = \dfrac{n^{1/n}-1}{n} \le \dfrac{2}{n^{3/2}}.$

Hence sum convergent.

Appended:

For $x \ge 0$, $n\ge 2 :$

$(1+x)^n \ge (n^2/4)x^2.$

(Proof: Use Binomial theorem.)

Set $x =\dfrac{2}{√n},$ then

$(1+x)^n = (1+2/√n)^n \ge$

$ (n^2/4)(4/n)= n.$

Hence: $1+2/√n \ge n^{1/n}$.

Answer 3

Note that

$$ \frac{n^{1/n} - 1}{n}\sim \frac{\log n}{n^2}$$

thus the given series converges by compartison with

$$\sum\limits_{n=1}^\infty \frac{1}{n^\frac32}$$

Answer 4

Let

$$t:=\frac{n^{1/n}-1}n$$ so that by the binomial theorem

$$(1+nt)^n=1+n^2t+\frac{n^3(n-1)}2t^2+R(t)=n$$ where $R(t)$ is positive. By solving the quadratic equation for the positive root, we can easily show that

$$t=O(n^{-3/2})$$ and this ensures convergence.