# Convergence of the series $\sum\limits_{n=1}^{\infty}\frac1n\log\left(1+\frac1n\right)$.

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.

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

• Oh, I didn't think of the comparison test. Thank you. – Arjun Banerjee Oct 14 '18 at 6:34
• You are welcome! – Michael Rozenberg Oct 14 '18 at 6:35
• Getting the exact value of $\sum_n \frac{1}{n^2}$ is harder than showing the sum converges – mathworker21 Oct 14 '18 at 6:57
• @mathworker21 Yes, but by the telescoping sum easy to see that the second sum converges. – Michael Rozenberg Oct 14 '18 at 7:45

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.”

• Thank you for clearing my doubts. But can't I avoid comparison test anyway? – Arjun Banerjee Oct 15 '18 at 4:45
• @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.) – Martin R Oct 15 '18 at 5:31

$$a_n:= (1/n)\log (1+1/n);$$

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 , and with

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

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

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

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