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 $$\displaystyle{\sum\limits_{n = 1}^{4N} {\frac{{i – {i^n}}}{n}} = i{H_{4N}} – \sum\limits_{n = 1}^{4N} {\frac{{{i^n}}}{n}} \approx i\left( {\ln 4N + \gamma } \right) + \ln \left( {1 – i} \right) = i\left( {2\ln 2 + \ln N + \gamma } \right) + \ln \left( {\sqrt 2 {e^{ – i\frac{\pi }{4}} }} \right) = i\left( {2\ln 2 + \ln N + \gamma } \right) – i\frac{\pi }{4} + \frac{1}{2}\ln 2}$$
Best Answer
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 $$ \Omega=\frac{3}{4} \ln2 - \frac{\ln \pi}{4}+\frac{\gamma}{4}-\frac{\pi}{8}$$