Does this series converge for any $1 \leq p < \infty$, $p \in \mathbb{R}$ Let $(a_n) = \left(1 - \frac{1}{n^{1/n}}\right)$.
Now consider the series $\, S = \sum_{n=0}^\infty |a_n|^p,\,  1 \leq p < \infty, \, p \in \mathbb{R}$. I am interested to know whether the series $S$ diverges for all $p$ or is there some $p$ for which the series converges. I am not able to prove that it diverges for all $p$ but I think that would be the case. I tried some of the basic convergence tests I know such as comparison test, ratio test, root test, integral test, Cauchy's condensation test.
Can someone help out here ?
 A: At least for $p > 2$ the series is convergent: Set $c_n:=n^{1/n}-1$. For $n\ge 2$ we have
$$
n=(c_n+1)^n=\sum_{k=0}^n {n \choose k} c_n^k \ge {n \choose 2}c_n^2=\frac{n(n-1)}{2}c_n^2,
$$
hence
$$
0 \le c_n \le \frac{\sqrt{2}}{\sqrt{n-1}},
$$
so
$$
0 \le a_n=\frac{c_n}{n^{1/n}} \le c_n \le \frac{\sqrt{2}}{\sqrt{n-1}}.
$$
For $p>2$ this yields a convergent majorant:
$$
a_n^p \le \frac{2^{p/2}}{(n-1)^{p/2}}.
$$
A: $\begin{array}\\
a_n
&= 1 - \frac{1}{n^{1/n}}\\
&= 1 - n^{-1/n}\\
&= 1 - e^{-\ln(n)/n}\\
&= 1 - (1-\frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2}))\\
&= \frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2})\\
&= \frac{\ln(n)}{n}(1+O(\frac{\ln(n)}{n}))\\
a_n^p
&= \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))^p\\
&= \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))\\
\sum_{n=1}^{\infty} a_n^p
&= \sum_{n=1}^{\infty} \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))\\
\end{array}
$
This converges for
$p > 1$
since
$\ln(n)
=o(n^c)
$
for any $c > 0$
(choose
$c$ such that
$cp < p-1$
such as
$c = (p-1)/(2p)$
so that
$\frac{\ln^p(n)}{n^p}
=o(\frac{n^{(p-1)/2}}{n^p})
=o(\frac1{n^{(p+1)/2}})
$
and
$(p+1)/2 > 1$)
and diverges for
$p \le 1$
by comparison with
$\sum \frac1{n}
$.
