Consider the series $\sum_{n=1}^{\infty }a_{n}$ where $a_{n}> 0$ for all $n\in \mathbb{N}$. Assume that: $\lim_{n \to \infty }n \ln\left ( \frac{a_{n}}{a_{n+1}} \right )=g$.

I need to prove that if $g> 1$, then $\sum_{n=1}^{\infty }a_{n}$ converges. Similarly, if $g< 1$, then $\sum_{n=1}^{\infty }a_{n}$ diverges.

Here is the solution to this problem as given to me, however I couldn't understand the last part of this proof. If someone has an idea, please share. In the solution to this problem, we need the following lemma (which I will not prove):

Lemma: Let $\sum_{n=1}^{\infty }a_{n}$ and $\sum_{n=1}^{\infty }b_{n}$ be two series of positive terms that satisfy the following inequality: $\frac{a_{n+1}}{a_{n}}\leq \frac{b_{n+1}}{b_{n}}$ for $n\geq K$. One can prove that if $\sum_{n=1}^{\infty }b_{n}$ converges, then $\sum_{n=1}^{\infty }a_{n}$ also converges.

If $g>1$ and let $\epsilon > 0$ be so small so that $g-\epsilon>1$. Then, for sufficiently large $n$: $n\ln\left ( \frac{a_{n}}{a_{n+1}} \right )>g-\epsilon$. Also, from the inequality: $\left ( 1+\frac{1}{n} \right )^{n}< e< \left ( 1+\frac{1}{n} \right )^{n+1}$, it follows that: $n\ln\left ( 1+\frac{1}{n} \right )<1$. So, $n\ln\left ( 1+\frac{1}{n} \right )<1<g-\epsilon <n\ln\left ( \frac{a_{n}}{a_{n+1}} \right ).$

I can understand everything up to this part. The part I couldn't get is the following: The solution says that it follows from the above inequality that $$\frac{a_{n+1}}{a_{n}}<\frac{\frac{1}{(n+1)^{g-\epsilon }}}{\frac{1}{n^{g-\epsilon }}}.\tag{*}\label{lab} $$

Then we use the previous lemma to conclude the convergence of the series. Can anyone tell me how to derive the inequality \eqref{lab}? Thanks.


Let $u=g-\epsilon$, then $n\log(a_n/a_{n+1})\gt u$, hence $a_{n+1}/a_n\lt\mathrm e^{-u/n}$. Since $\mathrm e\gt\left(1+\frac1n\right)^n$, this yields $\mathrm e^{-u/n}\lt\left(1+\frac1n\right)^{-u}$. The RHS is the odd-loking ratio $\frac{\frac1{(n+1)^u}}{\frac1{n^u}}$ in the RHS of $(*)$.

Edit The other case is $g\lt1$, then $n\log(a_n/a_{n+1})\lt1$ for every $n$ large enough, hence $a_{n+1}/a_n\gt\mathrm e^{-1/n}$. Since $\mathrm e\lt\left(1+\frac1{n-1}\right)^{n}$, $\mathrm e^{-1/n}\gt\frac{n-1}{n}$. Thus, the sequence $(na_{n+1})_n$ is nondecreasing for $n$ large enough. In particular, $a_{n+1}\gt C/n$ for every $n$ large enough, for some positive $C$, and this implies that the series $\sum\limits_na_n$ diverges.

  • $\begingroup$ Yeah, it worked. Thanks a lot. $\endgroup$
    – C. Lambda
    Jul 28 '12 at 16:22
  • $\begingroup$ The case of divergence follows from applying a similar argument $\endgroup$
    – C. Lambda
    Jul 28 '12 at 16:38
  • 1
    $\begingroup$ Indeed. Glad that you managed to see this by yourself... :-) $\endgroup$
    – Did
    Jul 28 '12 at 16:47

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