Digamma series

January 16, 2021January 16, 2021 Tolaso

Let $\psi^{(0)}$ denote the digamma series. Evaluate the series

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

Solution

The series telescopes;

$\begin{align*} \mathcal{S} &=\sum_{n=1}^{\infty} \bigg(\psi^{(0)} \left(\frac{1+2n-1}{2} \right) -\psi^{(0)} \left( \frac{2n-1}{2} \right) - \\ &\quad \quad \quad -\frac{1}{2n-1}+\psi^{(0)} \left( \frac{1+2n}{2}\right)- \psi^{(0)}\left(\frac{2n}{2} \right)-\frac{1}{2n} \bigg)\\ &=\sum_{n=1}^{\infty} \bigg( \psi^{(0)}\left( n \right)-\psi^{(0)} \left( n-\frac{1}{2} \right) -\frac{1}{2n-1}+ \\ &\quad \quad \quad +\psi^{(0)} \left( n+\frac{1}{2} \right)-\psi^{(0)} \left( n \right) -\frac{1}{2n} \bigg) \\ &=\sum_{n=1}^{\infty} \bigg( \psi^{(0)} \left(n+\frac{1}{2}\right) -\psi^{(0)} \left(n-\frac{1}{2}\right)- \\ &\quad \quad \quad -\frac{1}{2n-1}-\frac{1}{2n} \bigg)\\ &=\sum_{n=1}^{\infty} \left( \frac{2}{2n-1}-\frac{1}{2n-1}-\frac{1}{2n} \right) \\ &=\sum_{n=1}^{\infty} \left( \frac{1}{2n-1}-\frac{1}{2n} \right)\\ &=\ln 2 \end{align*}$

Harmonic function

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Application of extended binomial theorem

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An integral identity

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A Gudermannian integral

2 thoughts on “Digamma series”

Tolaso says:

January 16, 2021 at 10:45 pm

Another solution goes along these lines;

$\begin{align*} \mathcal{S} &= \sum_{n=1}^{\infty} \left( \psi^{(0)}\left(\frac{1+n}{2}\right)-\psi^{(0)}\left(\frac{n}{2}\right)-\frac{1}{n} \right) \\ &=\lim_{N \rightarrow +\infty} \left ( \psi^{(0)} \left ( \frac{N+1}{2} \right ) - \psi^{(0)} \left ( \frac{1}{2} \right ) - \mathcal{H}_N \right ) \\ &= \lim_{N \rightarrow +\infty} \left ( \ln \frac{N}{2} + 2 \ln 2 + \gamma - \ln N - \gamma + \mathcal{O} \left ( \frac{1}{N} \right ) \right ) \\ &= \ln 2 \end{align*}$

since $\psi^{(0)} \left ( \frac{1}{2} \right )= - 2\ln 2 -\gamma$ , $\mathcal{H}_n = \gamma + \ln n + \mathcal{O} \left ( \frac{1}{n} \right )$ and $\psi^{(0)}(x) = \ln x + \mathcal{O} \left ( \frac{1}{x} \right )$ .

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Tolaso says:

January 16, 2021 at 11:06 pm

Another interesting series is the following:

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

Proof: Let $\mathcal{S}_n$ denote the sum

$\mathcal{S}_n = \sum_{n=1}^{2N}(-1)^n\left(\psi^{(0)}\left(\frac{n+1}2\right)-\psi^{(0)}\left(\frac{n}2\right)-\frac{1}{n} \right)$

Hence,

$\begin{align*} \mathcal{S}_n &=-\ln 2 -\gamma+2 \sum_{n=1}^{N} \left( \psi^{(0)}\left(n+\frac{1}{2} \right)-\psi^{(0)}(n) \right)-\\ &\quad \quad \quad - \psi^{(0)}\left( N+\frac{1}{2} \right) \\ &=-\ln 2 -\gamma + 4\sum_{n=1}^{N} \sum_{k=0}^{\infty} \frac{1}{(2k+2n)(2k+2n+1)} - \\ &\quad \quad \quad -\psi^{(0)}\left(N+\frac{1}{2} \right)\\ &=-\ln 2 -\gamma + 4\sum_{n=1}^{N} \sum_{k=n}^{\infty} \frac{1}{2k(2k+1)}- \psi^{(0)}\left( N+\frac{1}{2} \right)\\ &=-\ln 2 -\gamma + 4\sum_{k=1}^{\infty} \sum_{n=1}^{\min(k,N)} \frac{1}{2k(2k+1)} - \psi^{(0)}\left(N+\frac{1}{2} \right)\\ &=-\ln 2 -\gamma + 4\left( \sum_{k=1}^{N} \frac{1}{2(2k+1)} +\sum_{k=N+1}^{\infty} \frac{N}{2k(2k+1)} \right) -\\ &\quad \quad \quad - \psi^{(0)}\left(N+\frac{1}{2} \right)\\ &=-\ln 2 -\gamma + 4\left( \frac{1}{2} \left( \mathcal{H} _{2N+1}-\frac{1}{2} \mathcal{H}_N-1 \right)+\frac{1}{4} + \mathcal{O} \left( \frac{1}{N} \right) \right)- \\ &\quad \quad \quad - \psi^{(0)}\left(N+\frac{1}{2} \right)\\ &=-\ln 2 -\gamma + \left(2\mathcal{H}_{2N+1}-\mathcal{H}_N - 1 \vphantom{\frac1N}\right)-\\ &\quad \quad \quad - \ln N +\mathcal{O} \left( \frac{1}{N} \right)\\ &=-\ln 2 -\gamma+ \bigg( 2(\gamma+\ln 2 +\ln N ) - \\ &\quad \quad \quad - (\gamma + \ln N)-1\vphantom{\frac1N} \bigg)-\ln N +\mathcal{O} \left( \frac{1}{N} \right)\\ &=\ln 2 -1+\mathcal{O} \left( \frac{1}{N} \right) \end{align*}$

since

$\begin{align*} \psi^{(0)} \left( \frac{n+1}{2} \right)- \psi^{(0)} \left(\frac{n}{2} \right) &=\sum_{k=0}^{\infty} \left( \frac{1}{k+\frac{n}{2}}-\frac{1}{k+\frac{n+1}{2}} \right)\\ &=2\sum_{k=0}^{\infty} \left( \frac{1}{2k+n}-\frac{1}{2k+n+1}\right)\\ &=2\sum_{k=0}^{\infty}\frac{1}{(2k+n)(2k+n+1)} \end{align*}$

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