Problem - M. Tetiva (Romania)
Evaluate
\[\sum_{n=1}^\infty\left(H_n-\ln(n)-\gamma-\frac{1}{2n}\right)\]where $H_n=\sum_{k=1}^n\frac{1}{k}$ and $\gamma$ is the Euler-Mascheroni constant.
Solution
Call the given sum S. We show that
\[S=\frac{1}{2}+\frac{\gamma}{2}-\frac{\ln(2\pi)}{2}.\]Fist we examine $A_m=\sum_{n=1}^m H_n$. We may write $A_m$ out as
\[\begin{align*} A_m &= \frac{1}{1} \\ &+\frac{1}{1}+\frac{1}{2} \\ &+\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \\ &+\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \\ &\;\,\vdots \\ &+\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{m} \\ &= \sum_{k=1}^m\frac{m-k+1}{k},\text{ by summing down columns} \\ &= (m+1)H_m-m.\end{align*}\]The partial sums of the given series are then
\[\begin{align*} S_m &= \sum_{n=1}^m\left(H_n-\ln(n)-\gamma-\frac{1}{2n}\right) \\ &= A_m-\ln(m!)-\gamma m-\frac{1}{2}H_m \\ &= \left(m+\frac{1}{2}\right)H_m-\ln(m!)-(1+\gamma)m. \\\end{align*}\]Asymptotically, $H_m=\gamma+\ln(m)+\frac{1}{2m}+O\left(\frac{1}{m^2}\right)$ and $m!=\sqrt{2\pi m}\left(\frac{m}{e}\right)^m\left(1+O\left(\frac{1}{m}\right)\right)$, so
\[\begin{align*} S &= \lim_{m\to\infty}S_m \\ &= \lim_{m\to\infty}\left(\frac{1}{2}+\frac{\gamma}{2}-\frac{\ln(2\pi)}{2}+O\left(\frac{1}{m}\right)\right) \\ &= \frac{1}{2}+\frac{\gamma}{2}-\frac{\ln(2\pi)}{2}.\end{align*}\]