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 Posted: Tue Jun 13, 2017 6:22 am
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Location: Ioannina, Greece
Calculating the series $\sum_{m=1}^{+\infty}\frac{(-1)^{m+1}}{m+n-1}$, I get the half of $\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)$. Can you suggest an elegant and simple formula for $\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)$ or $\psi\big(n+\frac{1}{2}\big)-\psi(n)$, where $\psi$ is the digamma function and $n$ is a positive integer $>2$.

P.S. I am aware of certain formulas about this, but not a satisfying one, if such exists.

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 Posted: Tue Jun 13, 2017 12:44 pm
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Joined: Sat Nov 07, 2015 6:12 pm
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Location: Larisa
I suppose you don't want Gauss digamma theorem to be used, e? , but rather a more elegant one.

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 Posted: Tue Jun 13, 2017 12:49 pm
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Joined: Sat Nov 07, 2015 6:12 pm
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Location: Larisa
Grigorios Kostakos wrote:
$\psi\big(n+\frac{1}{2}\big)-\psi(n)$,

Well,

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

and

$$\psi^{(0)}(n) = - \gamma +\sum_{k=1}^{n-1} \frac{1}{k}$$

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 Posted: Tue Jun 13, 2017 1:02 pm
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Joined: Mon Nov 09, 2015 1:36 am
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Location: Ioannina, Greece
Tolaso J Kos wrote:
I suppose you don't want Gauss digamma theorem to be used, e? , but rather a more elegant one.

Toiis, I seek something about the difference $\psi\big(\tfrac{n+1}{2}\big)-\psi\big(\tfrac{n}{2}\big)$ and the more elegant I find is
$$\displaystyle\psi\big(\tfrac{n+1}{2}\big)-\psi\big(\tfrac{n}{2}\big)=\mathop{\sum}\limits_{k=1}^{\frac{n}{2}}\frac{2}{2k-1}-\mathop{\sum}\limits_{k=1}^{\frac{n}{2}-1}\frac{1}{k}-2\log2\,.$$
Does exists something more simple? I doubt.

edit: we were posting the same minute!

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