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I am trying to check the convergence or divergence of the series $\displaystyle\sum_{n=1}^{\infty}\dfrac1n\log\left(1+\dfrac1n\right)$.

My attempt: for a finite $p$,\begin{align}\displaystyle\sum_{k=n}^{n+p}\dfrac1k\log\left(1+\dfrac1k\right)&\lt\dfrac1n\displaystyle\sum_{k=n}^{n+p}\log\left(1+\dfrac1k\right)\\&=\dfrac1n\log\large\Pi_{k=n}^{n+p}\left(\dfrac{k+1}{k}\right)\\&=\dfrac1n\log\left(1+\dfrac{p+1}{n}\right)\\&\lt\dfrac1n\log2,\text{ for large $n$ and $p$ is finite.}\\&\lt\varepsilon\end{align} Hence the series converges.

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Because $$\frac{\frac{1}{n}\ln(1+\frac{1}{n})}{\frac{1}{n^2}}\rightarrow1$$ and $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.$$

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  • $\begingroup$ Oh, I didn't think of the comparison test. Thank you. $\endgroup$ – Arjun Banerjee Oct 14 '18 at 6:34
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Oct 14 '18 at 6:35
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    $\begingroup$ Getting the exact value of $\sum_n \frac{1}{n^2}$ is harder than showing the sum converges $\endgroup$ – mathworker21 Oct 14 '18 at 6:57
  • $\begingroup$ @mathworker21 Yes, but by the telescoping sum easy to see that the second sum converges. $\endgroup$ – Michael Rozenberg Oct 14 '18 at 7:45
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Your argument is not correct. For the Cauchy criterion you have to show that for every $\varepsilon > 0$ there is an $N \in \Bbb N$ such that $$ \sum_{k=n}^{n+p}\dfrac1k\log\left(1+\dfrac1k\right) < \varepsilon $$ for all $n \ge N$ and all $p \ge 0$. So you can not “fix” $p$ and assume that $$ \dfrac1n\log\left(1+\dfrac{p+1}{n}\right)\lt\dfrac1n\log2 \, . $$


But using the “well-known” estimate $\log x \le x-1$ one gets $$ 0 \le \frac1n\log\left(1+\dfrac1n\right) \le \frac{1}{n^2} $$ and that implies the convergence by the “squeeze theorem.”

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  • $\begingroup$ Thank you for clearing my doubts. But can't I avoid comparison test anyway? $\endgroup$ – Arjun Banerjee Oct 15 '18 at 4:45
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    $\begingroup$ @ArjunBanerjee: Well, you could write $\sum_{k=n}^{n+p}\frac1k\log\left(1+\frac1k\right) \le \sum_{k=n}^{n+p} \frac{1}{k^2} < \sum_{k=n}^{n+p} \frac{1}{k(k-1)} = \sum_{k=n}^{n+p} \left( \frac 1{k-1} - \frac 1k \right) = \frac{1}{n-1} - \frac 1{n+p} < \frac{1}{n-1}$ if that is what you are looking for. (But that essentially repeats the proof that $\sum \frac{1}{n^2}$ is convergent.) $\endgroup$ – Martin R Oct 15 '18 at 5:31
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$a_n:= (1/n)\log (1+1/n);$

Recall : $\lim_{n \rightarrow \infty}(1+1/n)^n=e$.

Hence $(1+1/n)^n$ is bounded by a $M$, real, positive.

Then $(1+1/n)^n <M$, and with

$\log (1+1/n)^n < \log M <M$:

$a_n = (1/n^2) \log(1+1/n)^n <M/n^2$.

Comparison test: $M \sum 1/n^2$ converges.

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  • $\begingroup$ Thank you for answering. $\endgroup$ – Arjun Banerjee Oct 15 '18 at 4:30
  • $\begingroup$ Arjun.A pleasure. $\endgroup$ – Peter Szilas Oct 15 '18 at 5:44

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