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Does $\sum_{n=2}^\infty \frac{\ln(n)}{n (n - 1)}$ converge?

Wolfram alpha suggests that the series converges. But I don't know yet how to prove it.

Attempting to apply the root test I got a complicated limit that I don't know how to evaluate: $$ \limsup_{n\to\infty}\sqrt[n]{\left|\frac{\ln(n)}{n (n - 1)}\right|},\quad a_n=\frac{\ln(n)}{n (n - 1)}. $$

Attempting to apply the ratio test I got $$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=1, $$ which means that the ratio test is inconclusive in this case.

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    It helps for these problems to remember that for any positive constant $a$, for all sufficiently large $n$, we have $\color{blue}{\ln n \le n^a}$. Then choose a value of $a>0$ that will help for the problem at hand, e.g. by letting us use the comparison test. For example, $a=1/2$ will help here. – Minus One-Twelfth Jun 01 '19 at 12:56

3 Answers3

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Yes, in this case the ratio test and the root test are inconclusive. My hint: show that, eventually, $$\dfrac{\ln(n)}{n (n - 1)}\leq \frac{1}{n^{3/2}}.$$ What may we conclude?

P.S. We can replace the exponent $3/2$ with any real number in $(1,2)$.

Robert Z
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Since$$\lim_{n\to\infty}\frac{\frac{\log n}{n(n-1)}}{\frac{\log n}{n^2}}=1,$$your series converges if and only if $\displaystyle\sum_{n=2}^\infty\frac{\log n}{n^2}$ converges. Which it does, by the integral test. Just use the fact that$$\int_2^\infty\frac{\log x}{x^2}\,\mathrm dx=\lim_{M\to\infty}-\frac1M-\frac{\log M}M+\frac12+\frac{\log2}2=\frac12+\frac{\log2}2.$$

José Carlos Santos
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The actual value is in https://oeis.org/A131688 $$\sum_{n\ge 2} \frac{\log n}{n^2(1-1/n)} = \sum_{n\ge 2}\sum_{j\ge 0} \frac{\log n}{n^{2+j}} = -\sum_{j\ge 0}\zeta'(2+j)\approx 1.2577468869443$$ where $\zeta'$ is the first derivative of the Riemann zeta function.

R. J. Mathar
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