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

What I have done so far

Lemma: $$\displaystyle{\mathop {\lim }\limits_{N \to \infty } \left( {\sum\limits_{n = 1}^N {\frac{1}{{4N - 1}}} - \frac {{\ln N}}{4}} \right) = \frac{3}{4}\ln 2 + \frac{1}{4}\gamma - \frac{\pi }{8}}$$ because even though $S = \sum_{n = 1}^N \frac{1}{4N - 1}$

$$\sum_{n = 1}^{4N} \frac{i - i^n}{n} = \sum_{n = 1}^N \frac{i - i^{4n - 3}}{4n - 3} + \sum_{n = 1}^N \frac{i - i^{4n - 2}}{4n - 2} + \sum_{n = 1}^N \frac{i - i^{4n - 1}}{4n - 1} + \sum_{n = 1}^N \frac{i - i^{4n}}{4n} \\ \quad = \sum_{n = 1}^N \frac{i + 1}{4n - 2} + \sum_{n = 1}^N \frac{2i}{4n - 1} + \sum_{n = 1}^N \frac{i - 1}{4n} $$\

$$\displaystyle{ = \frac{{i + 1}}{2}\sum\limits_{n = 1}^N {\frac{1}{{2n - 1}}} + 2i \cdot S + \frac{ {i - 1}}{4}\sum\limits_{n = 1}^N {\frac{1}{n}} = \frac{{i + 1}}{2}\left( {1 + \ frac{1}{3} + \frac{1}{5}.. + \frac{1}{{2N - 1}}} \right) + 2i \cdot S + \frac{{i - 1}} {4}{H_N} = }$$

$$\displaystyle{ = \frac{{i + 1}}{2}\left( {{H_{2N}} - \frac{1}{2}{H_N}} \right) + 2i \cdot S + \frac {{i - 1}}{4}{H_N} \approx \frac{{i + 1}}{2}\left( {\ln 2N + \gamma - \frac{1}{2}\left( { \ln N + \gamma } \right)} \right) + 2i \cdot S + \frac{{i - 1}}{4}\left( {\ln N + \gamma } \right) \approx }$$

$$\displaystyle{ \approx \frac{{i + 1}}{2}\left( {\ln 2 + \frac{1}{2}\ln N + \frac{1}{2}\gamma } \right ) + 2i \cdot S + \frac{{i - 1}}{4}\left( {\ln N + \gamma } \right) \Rightarrow \boxed{\sum\limits_{n = 1}^{4N } {\frac{{i - {i^n}}}{n}} \approx \frac{{i + 1}}{2}\ln 2 + \frac{i}{2}\left( {\ ln N + \gamma } \right) + 2i \cdot S}}$$

We also have $$\displaystyle{\sum\limits_{n = 1}^{4N} {\frac{{i - {i^n}}}{n}} = i{H_{4N}} - \sum\limits_{n = 1}^{4N} {\frac{{{i^n}}}{n}} \approx i\left( {\ln 4N + \gamma } \right) + \ln \left( {1 - i} \right) = i\left( {2\ln 2 + \ln N + \gamma } \right) + \ln \left( {\sqrt 2 {e^{ - i\frac{\pi }{4}} }} \right) = i\left( {2\ln 2 + \ln N + \gamma } \right) - i\frac{\pi }{4} + \frac{1}{2}\ln 2}$$

Gary
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Martin.s
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  • You can try to convert the inner sum to an integral, or you may try something like this. – Gary Nov 20 '23 at 07:41
  • the closed form is : $$ \frac{3}{4} \ln2 -\frac{\ln\pi}{4}+\frac{\gamma}{4}-\frac{\pi}{8} $$ – Faoler Nov 23 '23 at 10:25
  • @Faoler how? Can you explain? – Martin.s Nov 23 '23 at 16:36
  • @Martin.s firstly I used : $$ \frac{1}{2}\psi \left(n+\frac{1}{2} \right)=\sum_{k=1}^n \frac{1}{2k-1}-\frac{\gamma}{2}-\ln2$$ then use the integral form of digamma function which known as : $$ \psi (n)=\ln(n)-\frac{1}{2n}-\int_0^\infty \left(\coth(x)-\frac{1}{x} \right)e^{-2nx} dx $$ – Faoler Nov 23 '23 at 19:43

1 Answers1

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I comment to use integral form to solve it which is too long method but I had simpler way , I'd like to note in this method we will use diverge series but its correct for some reasons.
lets denote your problem by $\Omega$ and use limit $$ \Omega =\lim_{x\to -1} \left(\sum_{n=1}^{\infty} x^n \sum_{k=1}^\infty \frac{1}{2k-1}-\frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{x+1}} \ln n-\left(\frac{\gamma}{2}+\ln2 \right)\sum_{n=1}^\infty x^n \right)$$ by using Cauchy product for the first series and dirichlet eta function for the second and get $$ \Omega =\lim_{x\to -1} \left(\left(\sum_{n=1}^{\infty}\frac{x^n}{2n-1} \right)\left(\sum_{n=0}^{\infty} x^n\right)-\frac{1}{2}\eta'(x+1)-\left(\frac{\gamma}{2}+\ln2 \right) \frac{x}{1-x} \right)$$ then $$ \Omega =\lim_{x\to -1} \left(-\sqrt{-x} \tan^{-1}(\sqrt{-x})\left(\frac{1}{1-x}\right)-\frac{1}{2}\eta'(x+1)-\left(\frac{\gamma}{2}+\ln2 \right) \frac{x}{1-x} \right)$$ $$ =-\frac{\pi}{4}\ \frac{1}{2}-\frac{1}{2}\left(\frac{1}{2} \ln \pi-\frac{1}{2} \ln2 \right)-\left(\frac{\gamma}{2}+\ln2 \right) \frac{-1}{2} $$ finally we get $$ \Omega=\frac{3}{4} \ln2 - \frac{\ln \pi}{4}+\frac{\gamma}{4}-\frac{\pi}{8}$$

Faoler
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