Compute $\sum (1+\frac12+\dotsb+\frac1n-\ln (n+\frac12)-\gamma)$ 
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
 A: The Euler-Mascharoni constant $γ$ is also the limit of
$$
a_n=\sum_{k=1}^n\frac1k-\ln(n+\tfrac12)
$$
since the difference $\ln(1+\tfrac1{2n})$ to the original definition converges to zero. So the terms in the sum converge to zero.
Set
$$
b_n=a_n-a_{n+1}=-\tfrac1{n+1}-\ln(1-\tfrac1{2(n+1)})+\ln(1+\tfrac1{2(n+1)})=\tfrac1{12(n+1)^3}+O(\tfrac1{(n+1)^5})
$$
then
$$
a_n-γ=(a_n-a_{n+1})+(a_{n+1}-γ)=...=\sum_{k=n}^\infty b_k
$$
so that
$$
\sum_{n=1}^∞(a_n-γ)=\sum_{1\le n\le k}b_k=\sum_{k=1}^∞k\,b_k=\sum_{k=0}^∞(k+1)\,b_k+(γ-\ln(2))
$$
and from here the convergence of the original follows series.

For the partial sums of that transformed series one gets
\begin{align}
b_0+2b_1+...+nb_{n-1}
&=-n+\sum_{k=1}^n\ln\frac{2k+1}{2k-1}\\
&=-n+\ln\frac{(2n+1)^n\,2^n\,n!}{(2n)!}
\end{align}
which can now be approximated using the Stirling formula, resulting in the exact value of the given series as
$$
\frac{(2n+1)^n\,2^n\,n!}{e^n(2n)!}
=\frac{(2n+1)^n\sqrt{2\pi(n+\theta_n)}}{\sqrt{2\pi(2n+\theta_{2n})}(2n)^n}
=\frac{(1+\tfrac1{2n})^n}{\sqrt{2+\frac1n\theta_{mix}}}
\xrightarrow[n\to\infty]{}\frac{\sqrt{e}}{\sqrt2}
$$
so that 
$$
\sum_{n=1}^\infty(a_n-γ)=\frac12-\frac32\ln2+γ
$$
A: 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,
$$
\lim_{n\to\infty}\sum_{k=1}^n\left(1+\frac12+\cdots+\frac1k-\log\left(k+\tfrac12\right)-\gamma\right)
=\gamma+\tfrac12-\tfrac32\log(2)
$$
A: Another method, based on the properties of the digamma function :

A: Perhaps this result will be of use:
$$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}-\ln(n)\rightarrow \gamma.$$
