Compute $\sum (1+\frac12+\dotsb+\frac1n-\ln (n+\frac12)-\gamma)$

Question

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

Answer 1

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.

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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 $$ \sum_{n=1}^\infty(a_n-γ)=\frac12-\frac32\ln2+γ $$

Answer 2

Another method, based on the properties of the digamma function :

Answer 3

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, $$ \lim_{n\to\infty}\sum_{k=1}^n\left(1+\frac12+\cdots+\frac1k-\log\left(k+\tfrac12\right)-\gamma\right) =\gamma+\tfrac12-\tfrac32\log(2) $$