Test for divergence of series I'm trying to solve exercise 2614.1 from Demidovich's famous book of exercises on Analysis. Having solved the bit about convergence, I'm now stuck in trying to prove that a series of terms $a_n\geq 0$ is divergent if there exists $N$ such that $(1-\sqrt[n]{a_n})\frac{n}{\log{n}} \leq 1$ for all $n>N$. I've been trying to find another divergent series with which to compare it, or to prove that $a_n$ does not tend to zero. Any help will be appreciated.
 A: I'll prove both parts.
Suppose that for some $N$ and some $p$,
$(1-\sqrt[n]{a_n})\frac{n}{\log{n}}\geq p>1$ for $n>N$. From here it
follows that it's enough to show the convergence of $\sum
(1-p\frac{\log{n}}{n})^n$. Since
$(1-p\frac{\log{n}}{n})^n=\exp(n\log(1-p\frac{\log{n}}{n}))$, I'll use
an order argument using the series for $\log(1-x)$ and $\exp{x}$.
These are
$\log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-...-\frac{x^n}{n} + o(x^n), x\to 0$,
$\exp{x}=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...+\frac{x^n}{n!} + o(x^n), x\to 0$.
Thus, $\log(1-p\frac{\log{n}}{n})=-p\frac{\log{n}}{n}-p^2\frac{(\log{n})^2}{n^2}+o(\frac{(\log{n})^2}{n^2})$,
for $\frac{\log{n}}{n}\to 0$, so
$n\log(1-p\frac{\log{n}}{n})=-p\log{n}-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})$.
Therefore
$\exp(n\log(1-p\frac{\log{n}}{n}))=\exp(-p\log{n}-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))$
$=\exp(-p\log{n})\exp(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))=\frac{1}{n^p}\exp(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))$
$=\frac{1}{n^p}\left(
1-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})+o\left(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})
\right) \right)$
$=\frac{1}{n^p}\left(
1-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}) \right)$.
So we have $(1-p\frac{\log{n}}{n})^n=\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}+o(\frac{(\log{n})^2}{n^{p+1}})$.
Dividing the LHS of the last equation by
$\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}$ and evaluating the
limit, we find that
$\lim_{n\to\infty}\frac{(1-p\frac{\log{n}}{n})^n}{\frac{1}{n^p}-p^2\frac{(\log{n})^2}{n^{p+1}}}=1$.
So the series of the terms in the denominator converges iff if the
series of the terms in the numerator does. But since $p>1$, we know
that $\sum\frac{1}{n^p}$ converges, and
$\sum\frac{(\log{n})^2}{n^{p+1}}$ converges too by comparing with the
former. Hence $\sum (1-p\frac{\log{n}}{n})^n$ converges and so does
$\sum a_n$.
On to the divergence part. It's enough to prove that $\sum
(1-\frac{\log{n}}{n})^n$ diverges. All steps above are still valid for
$p=1$, so $\lim_{n\to\infty}\frac{(1-\log{n}/n))^n}{1/n-(\log{n})^2/n^2}=1$.
But $\sum(\log{n})^2/n^2$ can be shown to converge, by comparison
with the series $\sum\frac{1}{n^\frac{3}{2}}$. This implies that the series in the
denominator diverges and so does that in the numerator, which is what
we wanted to prove.
Note that the key here involves manipulation of $o-$little.
this test is known in
Russian circles as "Zhame test" 
A: Try to show that $(1 - \frac{{\log n}}{n})^n $ is asymptotically equal to $b_n$, as $n \to \infty$, where 
$\sum\nolimits_{n = 1}^\infty  {b_n }$ is some very well known divergent series...
