Evaluate the infinite sum of ln(n)/n^2 for n from 1 to infinity.

Step-by-step solution

1. 1

   Establish convergence first. Compare with $\sum n^{-3/2}$: for large $n$, $\ln n<n^{1/2}$, so $\dfrac{\ln n}{n^2}<\dfrac{1}{n^{3/2}}$. Since $\sum n^{-3/2}$ converges ($p=\tfrac32>1$), the given series converges absolutely. The $n=1$ term is $\ln 1=0$, so the sum effectively starts at $n=2$.

2. 2

   Differentiate the zeta function under the sum. For $\Re(s)>1$,

   $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$

   and $\dfrac{d}{ds}n^{-s}=\dfrac{d}{ds}e^{-s\ln n}=-\ln n\cdot n^{-s}$. Term-by-term differentiation is legitimate here because the differentiated series converges uniformly on $\Re(s)\ge1+\delta$.

3. 3

   Read off the identity. Therefore

   $\zeta'(s)=-\sum_{n=1}^{\infty}\frac{\ln n}{n^s}\quad\Longrightarrow\quad\sum_{n=1}^{\infty}\frac{\ln n}{n^s}=-\zeta'(s)$

   Setting $s=2$ gives the closed form $-\zeta'(2)$. There is no elementary expression: $\zeta'(2)$ relates to the Glaisher-Kinkelin constant, not to $\pi$ and $\ln 2$ alone.

4. 4

   Get a numerical value that is actually trustworthy. Direct summation converges far too slowly (the tail past $N$ is about $(\ln N+1)/N$). Adding the first $200{,}000$ terms and correcting the tail with the Euler-Maclaurin estimate $\dfrac{\ln N+1}{N}-\dfrac{f(N)}{2}-\dfrac{f'(N)}{12}$ gives

   $\sum_{n=1}^{\infty}\frac{\ln n}{n^2}=0.9375482543$

5. 5

   Check the magnitude independently. The known value $\zeta'(2)=-0.93754825431584\ldots$ agrees to ten digits with the computed sum ✓. A crude sanity bound also holds: the first four nonzero terms alone give $\tfrac{\ln2}{4}+\tfrac{\ln3}{9}+\tfrac{\ln4}{16}+\tfrac{\ln5}{25}=0.4464$, and the remaining positive tail carries the sum up toward $0.9375$ ✓.