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The taylor series expansion of the function $$f(x)=\ln(1+x)$$ around zero is: $$f(x)=\sum_{k=1}^\infty\dfrac{(-1)^{(k+1)}}{k}x^k$$ Putting $x=1$ we have the alternating series: $f(1)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...=\ln(2)$ The partial sum: $$S(N)=\sum_{k=1}^N\dfrac{(-1)^{(k+1)}}{k}$$ gives the result: $$S(N)=\ln(2)-\dfrac{1}{2}(-1)^N\left(\Psi\left(\dfrac{1}{2}N+1\right)-\Psi\left(\dfrac{1}{2}N+\dfrac{1}{2}\right)\right)$$ where $\Psi$ is the digamma function. So we have: $\displaystyle\lim_{N=+\infty}S(N)=\ln(2)$ which means: $$\displaystyle\lim_{N=+\infty}\left[-\dfrac{1}{2}(-1)^N\left(\Psi\left(\dfrac{1}{2}N+1\right)-\Psi\left(\dfrac{1}{2}N+\dfrac{1}{2}\right)\right)\right]=0$$ How is it possible to prove the last limit? Thanks.

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    $\begingroup$ The integral representation of the digamma function yields $$\psi(s+\tfrac12)-\psi(s)=\int_0^\infty\frac{e^{-st}-e^{-(s+1/2)t}}{1-e^{-t}}dt=\int_0^\infty \frac{e^{-st}}{1+e^{-t/2}}dt\leqslant\int_0^\infty e^{-st}dt=\frac1s.$$ More quantitatively, $$\lim_{s\to\infty}s\,(\psi(s+a)-\psi(s))=\frac1a.$$ $\endgroup$ – Did May 25 '16 at 15:34
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We have the identity: $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1} $$ hence by taking $a=\frac{N}{2}+1$ and $b=\frac{N}{2}+\frac{1}{2}$ it follows that: $$ \psi\left(\frac{N}{2}+1\right)-\psi\left(\frac{N}{2}+\frac{1}{2}\right)=2\sum_{n\geq 0}\frac{1}{(2n+N+2)(2n+N+1)}\tag{2}$$ but $\sum_{m\geq 0}\frac{1}{(2m+1)(2m+2)}$ is trivially converging, so the limit of $(2)$ as $N\to +\infty$ is simply zero.

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