# Computing $\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$

I'm sure you can do this easily, but I'm looking for an easy way that only uses series manipulation. Is that possible?

$$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$$

where $\psi^{(0)}(x)$ is digamma function

Here is a supplementary question, the alternating version

$$\sum_{n=1}^{\infty} (-1)^{n+1} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)$$ And this one will take into account our year as a power $$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)^{2014}$$

• What's $\psi^{(0)}$ here? – Simon S Nov 4 '14 at 13:10
• @SimonS I added the details. – user 1357113 Nov 4 '14 at 13:12
• Were you investigating the asymptotics of $~2^{-N}\cdot\displaystyle\prod_{n=1}^N \frac{\Gamma\bigg(\dfrac{n+1}2\bigg)}{\Gamma\bigg(\dfrac n2+1\bigg)}$ ? – Lucian Nov 4 '14 at 19:18
• @SimonS: $\displaystyle\psi^{(0)}(x)=\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^0\psi(x)=\psi(x)$. Don't ask me why; I'm just answering :-) – robjohn Nov 4 '14 at 20:52

You can calculate the series by combining the even and the odd summands. This yields:

$$\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n}\right)=\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(\frac{1+2n-1}{2}\right)-\psi^{(0)}\left(\frac{2n-1}{2}\right)-\frac{1}{2n-1}+\psi^{(0)}\left(\frac{1+2n}{2}\right)-\psi^{(0)}\left(\frac{2n}{2}\right)-\frac{1}{2n}\right)=\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(n\right)-\psi^{(0)}\left(n-\frac{1}{2}\right)-\frac{1}{2n-1}+\psi^{(0)}\left(n+\frac{1}{2}\right)-\psi^{(0)}\left(n\right)-\frac{1}{2n}\right)=\sum_{n=1}^{\infty} \left(\psi^{(0)}\left(n+\frac{1}{2}\right)-\psi^{(0)}\left(n-\frac{1}{2}\right)-\frac{1}{2n-1}-\frac{1}{2n}\right)=\sum_{n=1}^{\infty} \left(\frac{2}{2n-1}-\frac{1}{2n-1}-\frac{1}{2n}\right)=\sum_{n=1}^{\infty} \left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\ln(2)$$ The last result follows form the series expansion of $\ln(1+x)$.

• You're welcome! :) – Redundant Aunt Nov 4 '14 at 18:18

Main Question

We can telescope the first sum: \begin{align} \lim_{N\to\infty}\sum_{n=1}^N\left[\psi\left(\frac{n+1}2\right)-\psi\left(\frac{n}2\right)-\frac1n\right] &=\lim_{N\to\infty}\left[\psi\left(\frac{N+1}2\right)-\psi\left(\frac12\right)-H_N\right] \end{align} In this answer it is shown that $H_{-1/2}=-2\log(2)$; therefore, $\psi\left(\frac12\right)=-2\log(2)-\gamma$.

Asymptotically, $H_n=\gamma+\log(n)+O\left(\frac1n\right)$; therefore, $\psi(x)=\log(x)+O\left(\frac1x\right)$.

Thus, \begin{align} &\lim_{N\to\infty}\left[\psi\left(\frac{N+1}2\right)-\psi\left(\frac12\right)-H_N\right]\\ &=\lim_{N\to\infty}\left[\log\left(\frac{N}2\right)+2\log(2)+\gamma-\log(N)-\gamma+O\left(\frac1n\right)\right]\\[4pt] &=\log(2) \end{align}

A Formula For Later Use

Starting with \begin{align} \psi(x) &=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+x}\right)\tag{1}\\ &=-\gamma+H_{x-1}\tag{2} \end{align} we get \begin{align} \psi\left(\frac{n+1}2\right)-\psi\left(\frac{n}2\right) &=\sum_{k=0}^\infty\left(\frac1{k+\frac{n}2}-\frac1{k+\frac{n+1}2}\right)\\ &=2\sum_{k=0}^\infty\left(\frac1{2k+n}-\frac1{2k+n+1}\right)\\ &=2\sum_{k=0}^\infty\frac1{(2k+n)(2k+n+1)}\tag{3} \end{align} Using $n=1$ and $\psi(1)=-\gamma$ in $(3)$, we get $$-\gamma-\psi\left(\frac12\right)=2\log(2)\tag{4}$$ which is an alternate proof that $\psi\left(\frac12\right)=-2\log(2)-\gamma$.

Supplementary Question

We can apply $(3)$ to \begin{align} &\sum_{n=1}^{2N}(-1)^n\left[\psi\left(\frac{n+1}2\right)-\psi\left(\frac{n}2\right)-\frac1n\right]\\ &=-\log(2)-\gamma+2\sum_{n=1}^N\left[\psi\left(n+\frac12\right)-\psi(n)\right]-\psi\left(N+\frac12\right)\\ &=-\log(2)-\gamma+4\sum_{n=1}^N\sum_{k=0}^\infty\frac1{(2k+2n)(2k+2n+1)}-\psi\left(N+\frac12\right)\\ &=-\log(2)-\gamma+4\sum_{n=1}^N\sum_{k=n}^\infty\frac1{2k(2k+1)}-\psi\left(N+\frac12\right)\\ &=-\log(2)-\gamma+4\sum_{k=1}^\infty\sum_{n=1}^{\min(k,N)}\frac1{2k(2k+1)}-\psi\left(N+\frac12\right)\\ &=-\log(2)-\gamma+4\left[\sum_{k=1}^N\frac1{2(2k+1)}+\sum_{k=N+1}^\infty\frac{N}{2k(2k+1)}\right]-\psi\left(N+\frac12\right)\\ &=-\log(2)-\gamma+4\left[\frac12\left(H_{2N+1}-\frac12H_N-1\right)+\frac14+O\left(\frac1N\right)\right]-\psi\left(N+\frac12\right)\\ &=-\log(2)-\gamma+\left[2H_{2N+1}-H_N-1\vphantom{\frac1N}\right]-\log(N)+O\left(\frac1N\right)\\ &=-\log(2)-\gamma+\left[2(\gamma+\log(2)+\log(N))-(\gamma+\log(N))-1\vphantom{\frac1N}\right]-\log(N)+O\left(\frac1N\right)\\ &=\log(2)-1+O\left(\frac1N\right) \end{align} Thus, changing the sign and taking the limit, we get $$\sum_{n=1}^\infty(-1)^{n+1}\left[\psi\left(\frac{n+1}2\right)-\psi\left(\frac{n}2\right)-\frac1n\right] =1-\log(2)$$

• Very good! (+1) – user 1357113 Nov 5 '14 at 8:14