Convergence of $\sum_{n=1}^{\infty}\left( \sqrt[n]{n}-1 \right)^p$

Question

I am looking for a way to conclude about the convergence of

$$\sum_{n=1}^{\infty}\left( \sqrt[n]{n}-1 \right)^p$$ depending on $p$. Do you have any clues as to what convergence test might be most applicable here?

Answer 1

It is immediate to notice that if $p<0$ then the term of the series tends to $\infty$ (because it is known that $\sqrt[n] n \to 1$), so in this case the series diverges. If $p=0$ then again the series diverges trivially, so let us study the case $p>0$.

First, let us find another series having the same behaviour as the given one. Notice that if $x = \frac 1 n$ then

$$\sqrt[n] n -1 = x^{-x} - 1 = \Bbb e ^{-x \log x} - 1 = \sum _{k \ge 0} \frac {(-x \log x)^k} {k!} - 1 = - x \log x + \dots = \frac {\log n} n + \dots .$$

This gives us the idea of comparing the term $\sqrt[n] n -1$ with $\frac {\log n} n$. Let us do it properly:

$$\lim _{n \to \infty} \frac {\sqrt[n] n -1} {\frac {\log n} n} = \left[ \text{let } x = \frac 1 n \right] = \lim _{x \to 0, x > 0} \frac {\Bbb e ^{-x \log x} - 1} {-x \log x} = \\ [\text{let } t = -x \log x \text{ and notice that } \lim _{x \to 0, x>0} -x \log x = 0 \text{, so } t \to 0] = \\ \lim _{t \to 0} \frac {\Bbb e^t - 1} t = (\Bbb e^t)' (0) = 1$$

so indeed the given series and the series $\sum _{n \ge 1} \left( \frac {\log n} n \right)^p$ have the same behaviour (by the limit comparison test), therefore it will be sufficient to study the convergence of the new series. We shall do this with the aid of Cauchy's condensation test. In order to use it, you'll have to show that the function $x \mapsto \frac {\log x} x$ decreases to $0$, but I'll leave this to you (just study the derivative of this function, it's easy). Cauchy's condensation test will then tell us that our series has the same behaviour as the series

$$\sum _{n \ge 1} 2^n \left( \frac {\log 2^n} {2^n} \right)^p = (\log^p 2) \sum _{n \ge 1} 2^n \frac {n^p} {2^{np}} = (\log^p 2) \sum _{n \ge 1} \frac {n^p} {2^{n(p-1)}} .$$

This last series is easy to attack now, using the root test:

$$\lim _{n \to \infty} \sqrt[n] {\frac {n^p} {2^{n(p-1)}}} = \lim _{n \to \infty} \frac {\sqrt[n] n ^p} {2^{p-1}} = \frac 1 {2^{p-1}} .$$

The root test tells us that the last series converges when $\frac 1 {2^{p-1}} < 1$ (which means $p>1$), that it diverges when $\frac 1 {2^{p-1}} > 1$ (which means $p<1$), and it isn't able to tell us anything when $\frac 1 {2^{p-1}} = 1$ (which means $p=1$).

Fortunately, when $p=1$ the last series becomes just $(\log 2) \sum _{n \ge 1} n$, which is easily seen to diverge.

Collecting everything we have obtained, we conclude that the given series converges if and only if $p>1$.

Answer 2

Hint. One may write, as $n \to \infty$, $$ \begin{align} \left( \sqrt[n]{n}-1 \right)^p&=\left( e^{\large \frac{\ln n}n}-1 \right)^p \\\\&=\left(\frac{\ln n}n+ O\left(\frac{\ln^2 n}{n^2}\right)\right)^p \\\\&=\frac{\ln^p n}{n^p}\cdot\left(1+ O\left(\frac{\ln n}{n}\right)\right)^p \\\\&=\frac{\ln^p n}{n^p}+ O\left(\frac{\ln^{p+1} n}{n^{p+1}}\right) \, \, (\text{by the binomial formula}) \end{align} $$ then, by comparison, the given series is convergent if and only if $p>1$.