could anyone help me figure out whether this infinite series $$\sum_{n=1}^{\infty }\left ( 1-\frac{\ln(n)}{n} \right )^{2n}$$ diverges?

I've tried using Cauchy's and d'Alembert's limit tests but both gave the result 1. I've also tried the necessary condition for convergence, but $$\lim_{n\rightarrow \infty }\left ( 1-\frac{\ln(n)}{n} \right )^{2n}=0$$

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    $\begingroup$ You can use $1+x \leqslant e^{x}$. $\endgroup$ – Daniel Fischer May 19 '14 at 15:24


$$\sum_{n=1}^\infty \Bigl( 1 - \frac{\log n}{n} \Bigr)^{2n} \leq \sum_{n=1}^\infty \biggl(\exp\Bigl( -\frac{\log n}{n} \Bigr)\biggr)^{2n} =\sum_{n=1}^\infty \exp( - 2\log n ) = \sum_{n=1}^\infty \frac{1}{n^2} < \infty.$$

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    $\begingroup$ Thank you! Really elegant and simple, I don't understand how I could have missed that. $\endgroup$ – label May 19 '14 at 16:57

Using Cauchy Condensation, we get the series. $$\sum_{n=1}^\infty 2^nf(2^n) = \sum_{n=1}^{\infty}\left( 1 - \frac{n \log 2}{2^n}\right)^{2^{n+1}} 2^n \tag{1}$$ Here we can use root test, $$\lim_{n\to\infty} \sqrt[n]{\left( 1 - \frac{n \log 2}{2^n}\right)^{2^{n+1}} 2^n} = \frac 1 2 \tag{2}$$ to argue that $(1)$ converges which implies the convergence of out original series. To evaluate the limit $(2)$, $$2 \lim_{n\to\infty} \left( 1 - \frac{n \log 2}{2^n}\right)^{\frac{2^n}{n \log 2}2 \log 2} = 2 e^{- 2\log 2} = \frac 1 2$$

  • $\begingroup$ It seems that your method of thinking like mine;-) $\endgroup$ – user63181 May 19 '14 at 15:36
  • $\begingroup$ @user63181 yep!! I was wondering if you missed $2n$ for $2^n$ down in denominator?? $\endgroup$ – Santosh Linkha May 19 '14 at 15:37
  • $\begingroup$ Thank you very much. Also for making me familiar with Cauchy Condensation. $\endgroup$ – label May 19 '14 at 16:59
  • $\begingroup$ @Dominik you are welcome :) usually Cauchy Condensation is weapon of choice for $\log$ $\endgroup$ – Santosh Linkha May 19 '14 at 17:00

This is convergent.

Hint (not a concrete proof): $$ (1-\frac{\ln n}{n})^{\frac{n}{\ln n} 2\ln n}\sim e^{-2\ln n}=\frac{1}{n^2}. $$


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