TITLE: Compute $\sum (1+\frac12+\dotsb+\frac1n-\ln (n+\frac12)-\gamma)$
QUESTION [5 upvotes]: Compute
$$\sum_{n=1}^\infty \Big(1+\frac12+\dotsb+\frac1n-\ln (n+\frac12)-\gamma\Big)$$
where $\gamma$ is Euler's constant


It seems to be difficult, I have no idea go get started
Thank you very much

REPLY [3 votes]: Using simple series rearrangement:
$$
\begin{align}
&\sum_{k=1}^n\left(1+\frac12+\cdots+\frac1k-\log\left(k+\tfrac12\right)-\gamma\right)\\
&=\sum_{k=1}^n\left(\sum_{j=1}^k\frac1j-\log\left(k+\tfrac12\right)-\gamma\right)\\
&=\sum_{j=1}^n\sum_{k=j}^n\frac1j-\sum_{k=1}^n\log\left(k+\tfrac12\right)-n\gamma\\
&=\sum_{j=1}^n\frac{n-j+1}j-\sum_{k=1}^n\log(2k+1)-n(\gamma-\log(2))\\
&=(n+1)H_n-\log\left(\frac{(2n+1)!}{2^nn!}\right)-n(\gamma-\log(2)+1)\\
&=\color{#C00000}{(n+1)H_n}\color{#00A000}{-\log\left(\frac{(2n+1)!}{n!}\right)}-n(\gamma-2\log(2)+1)\\[6pt]
&=\color{#C00000}{(n+1)(\log(n)+\gamma+\tfrac1{2n})}\\
&\color{#00A000}{-(2n+1)\log(2n+1)+(2n+1)-\tfrac12\log(2\pi(2n+1))}\\
&\color{#00A000}{+n\log(n)-n+\tfrac12\log(2\pi n)}\\
&-n(\gamma-2\log(2)+1)+o(1)\\[12pt]
&=\gamma+\tfrac12-\tfrac32\log(2)+o(1)\\
\end{align}
$$
In red, we used the asymptotic expansion for $H_n$, and in green, we used Stirling's formula.
Therefore,
$$
\lim_{n\to\infty}\sum_{k=1}^n\left(1+\frac12+\cdots+\frac1k-\log\left(k+\tfrac12\right)-\gamma\right)
=\gamma+\tfrac12-\tfrac32\log(2)
$$