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I apologise if this is a dumb question, but I have trouble deriving $\displaystyle\psi\left(\frac{1}{2}\right)=-\gamma-2\ln{2}$. I have tried the following. \begin{align} \psi\left(\frac{1}{2}\right) &=\lim_{N\to\infty}\left[-\gamma-2+\sum^N_{k=1}\left(\frac{1}{k}-\frac{1}{k+1/2}\right)\right]\\ &=\lim_{N\to\infty}\left[-\gamma-1+\sum^{\lfloor \frac{N-1}{2} \rfloor}_{k=1}\frac{1}{2k}+\sum^{\lfloor \frac{N-1}{2} \rfloor}_{k=1}\frac{1}{2k+1}-2\sum^N_{k=1}\frac{1}{2k+1}\right]\\ &=\lim_{N\to\infty}\left[-\gamma-1+\sum^{\lfloor \frac{N-1}{2} \rfloor}_{k=1}\left(\frac{1}{2k}-\frac{1}{2k+1}\right)-2\sum^N_{k=1+\lfloor \frac{N-1}{2} \rfloor}\frac{1}{2k+1}\right]\\ &=-\gamma-\ln{2}-\lim_{N\to\infty}\sum^N_{k=1+\lfloor \frac{N-1}{2} \rfloor}\frac{2}{2k+1} \end{align} I have trouble showing that the last sum $\to \ln{2}$ as $N \to \infty$, and would like to seek your assistance in evaluating the limit. Thank you in advance.

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Note the simple fact that

$$\lim_{N\to \infty} \sum_{k=1+\lfloor{\frac{N-1}{2}}\rfloor}^{N} \frac{2}{2k+1} =\lim_{n\to\infty} \int_{n}^{2n} \frac{2}{2x+1} \ dx= \log(2)$$

Q.E.D.

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One could evaluate the limit as follows. Set $N=2 n$: $$\lim_{N\to \infty} \sum_{k=1+\lfloor{\frac{N-1}{2}}\rfloor}^{N} \frac{2}{2k+1}=\lim_{n\to \infty} \sum_{k=n}^{2n} \frac{2}{2k+1}=\lim_{n\to \infty} 2\cdot\left(\sum_{k=0}^{2n} \frac{1}{2k+1}-\sum_{k=0}^{n-1} \frac{1}{2k+1}\right)=2\cdot\lim_{n\to \infty} \left(\left(\sum_{k=1}^{4n+1} \frac{1}{k}-\sum_{k=1}^{2n} \frac{1}{2k}\right)-\left(\sum_{k=1}^{2n-1} \frac{1}{k}-\sum_{k=1}^{n-1} \frac{1}{2k}\right)\right)=2\cdot\lim_{n\to \infty}{H_{4n+1}-\frac{1}{2}H_{2n}-H_{2n-1}+\frac{1}{2}H_{n-1}}=2\cdot\lim_{n\to \infty}{(H_{4n+1}-\ln(4n+1))+\ln(4n+1)-\frac{1}{2}(H_{2n}-\ln(2n))-\frac{1}{2}\ln(2n)-(H_{2n-1}-\ln(2n-1))-\ln(2n-1)+\frac{1}{2}(H_{n-1}-\ln(n-1))+\frac{1}{2}\ln(n-1)}=2\cdot\lim_{n\to \infty}{(H_{4n+1}-\ln(4n+1))-\frac{1}{2}(H_{2n}-\ln(2n))-(H_{2n-1}-\ln(2n-1))+\frac{1}{2}(H_{n-1}-\ln(n-1))}+2\cdot\lim_{n\to \infty}{\ln(4n+1)-\frac{1}{2}\ln(2n)-\ln(2n-1)+\frac{1}{2}\ln(n-1)}=2\cdot(\gamma-\frac{1}{2}\gamma-\gamma+\frac{1}{2}\gamma)+2\cdot\lim_{n\to \infty}{\ln(\frac{4n+1}{2n-1})-\frac{1}{2}\ln(\frac{2n}{n-1})}=2\cdot(\ln(2)-\frac{1}{2}\ln(2))=\ln(2)$$ Here I used that: $$\lim_{n\to \infty}H_n-\ln(n)=\gamma$$ Where $\gamma$ is Euler's constant. Note that there is a faster way to evaluate $\psi\left(\frac{1}{2}\right)$: $$\psi\left(\frac{1}{2}\right)=-\gamma-2+\sum_{k=1}^{\infty} \left(\frac{1}{k}-\frac{1}{k+\frac{1}{2}}\right)=-\gamma+\sum_{k=1}^{\infty} \left(\frac{1}{k}-\frac{1}{k-\frac{1}{2}}\right)=-\gamma+\sum_{k=1}^{\infty} \left(\frac{2}{2k}-\frac{2}{2k-1}\right)=-\gamma-2\cdot\sum_{k=1}^{\infty} \left(\frac{1}{2k-1}-\frac{1}{2k}\right)=-\gamma-2\cdot\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=-\gamma-2\ln(2)$$ I hope, this is helpful and I apologize for all mistakes in my English, I'm not a native speaker. :)

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  • $\begingroup$ Thanks, this was quite helpful and I feel rather stupid now. $\endgroup$ – SuperAbound Sep 7 '14 at 12:04
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

See $\ds{\bf 6.3.22}$ in this table.

\begin{align} \color{#66f}{\large\Psi\pars{\half}+ \gamma}& =\int_{0}^{1}{1 - t^{-1/2} \over 1 - t}\,\dd t\ =\ \overbrace{-\int_{0}^{1}{1 - t^{1/2} \over t^{1/2}\pars{1 - t}}\,\dd t} ^{\ds{\color{#c00000}{t^{1/2}\ \equiv\ x\ \imp\ t\ =\ x^{2}}}}\ =\ -\int_{0}^{1}{1 - x \over x\pars{1 - x^{2}}}\,2x\,\dd x \\[5mm]&=-2\int_{0}^{1}{\dd x \over 1 + x} =\color{#66f}{\large -2\ln\pars{2}} \end{align}

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As shown in this answer, $H_{-1/2}=-2\log(2)$. Therefore, $$ \begin{align} \psi(1/2) &=H_{-1/2}-\gamma\\ &=-2\log(2)-\gamma \end{align} $$

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