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Digamma and Trigamma functions

Let \psi^{(0)} and \psi^{(1)} denote the digamma and trigamma functions respectively. Prove that:

    \[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n + \frac{1}{2} \psi^{(1)}(n) \right ) = 1+ \frac{\gamma}{2} - \frac{\ln 2\pi}{2}\]

where \gamma denotes the Euler – Mascheroni constant.

Solution

We begin with the recently discovered identity:

    \begin{align*} \log \Gamma(1+x) &= \frac{\ln 2 \pi -1}{2} -\gamma \left ( x+\frac{1}{2} \right ) \\ &\quad + \frac{2x-1}{2} + \sum_{n=1}^{\infty} \left ( \psi^{(0)}(n+1) -\ln(x+n) +\frac{2x-1}{2(1+n)} \right ) \end{align*}

Letting x=1 we get that

    \[\sum_{n=1}^{\infty} \left ( \psi^{(0)}(n) - \ln n +\frac{1}{2n} \right ) = \frac{1+\gamma-\ln 2\pi}{2}\]

Now combining this result here we conclude the exercise.

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