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.

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    $\begingroup$ The condition is equivalent to a condition of the form a_n \ge (something). So the problem statement is equivalent to showing that \sum (something) diverges. Write down (something). $\endgroup$ – Qiaochu Yuan Dec 10 '10 at 1:34
  • $\begingroup$ Thank you. That something is $(1-\log{n}/n)^n$. Root test here is inconclusive. Can you give me a hint for the divergence of this? $\endgroup$ – Weltschmerz Dec 10 '10 at 1:45
  • $\begingroup$ What is the limit of (1 - x/n)^n as n goes to infinity? Can you adapt the argument to the case where x varies slowly with n? $\endgroup$ – Qiaochu Yuan Dec 10 '10 at 2:02
  • $\begingroup$ Sorry remove my answer, but something was missing for a complete solution $\endgroup$ – Bryan Yocks Dec 10 '10 at 2:09

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 $\exp(n\log(1-p\frac{\log{n}}{n}))=\exp(-p\log{n}-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}))$


$=\frac{1}{n^p}\left( 1-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n})+o\left(-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}) \right) \right)$

$=\frac{1}{n^p}\left( 1-p^2\frac{(\log{n})^2}{n}+o(\frac{(\log{n})^2}{n}) \right)$.

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"

  • $\begingroup$ People actually use this test? $\endgroup$ – Qiaochu Yuan Dec 20 '10 at 20:12

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...


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