Find the value Sumation

Question

We have to find the value of $$ \sum_{k=0}^\infty \left[ \frac{3\ln(4k+2)}{4k+2} -\frac{\ln(4k+3)}{4k+3} -\frac{\ln(4k+4)}{4k+4} -\frac{\ln(4k+5)}{4k+5} \right] $$

My try:

But I am stuck after that.

Answer 1

The Euler-Maclaurin Sum Formula says $$ \sum_{k=1}^n\frac{\log(k)}{k}=\frac12\log(n)^2+C+O\!\left(\frac{\log(n)}{n}\right) $$

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Therefore, $$ \begin{align} \sum_{k=1}^{2n}(-1)^k\frac{\log(k)}k &=2\sum_{k=1}^n\frac{\log(2k)}{2k}-\sum_{k=1}^{2n}\frac{\log(k)}k\\ &=\sum_{k=1}^n\frac{\log(2)+\log(k)}k-\sum_{k=1}^{2n}\frac{\log(k)}k\\ &=\frac12\log(n)^2+\log(2)\log(n)+\gamma\log(2)-\frac12\log(2n)^2+O\!\left(\frac{\log(n)}{n}\right)\\[9pt] &\stackrel{n\to\infty}\to\gamma\log(2)-\frac12\log(2)^2 \end{align} $$ and $$ \begin{align} 2\sum_{k=1}^{2n}(-1)^k\frac{\log(2k)}{2k} &=4\sum_{k=1}^n\frac{\log(4k)}{4k}-2\sum_{k=1}^{2n}\frac{\log(2k)}{2k}\\ &=\sum_{k=1}^n\frac{\log(4)+\log(k)}k-\sum_{k=1}^{2n}\frac{\log(2)+\log(k)}k\\ &=\frac12\log(n)^2+\log(4)\log(n)+\gamma\log(4)\\ &-\frac12\log(2n)^2-\log(2)\log(2n)-\gamma\log(2)+O\!\left(\frac{\log(n)}{n}\right)\\[9pt] &\stackrel{n\to\infty}\to\gamma\log(2)-\frac32\log(2)^2 \end{align} $$

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After computing the sums above, it turns out that the sum in the question is a bit simpler $$ \begin{align} \overbrace{\sum_{k=1}^\infty(-1)^k\frac{\log(k)}k}^{\substack{+1\leftrightarrow0\pmod2\\-1\leftrightarrow1\pmod2}} -\overbrace{2\sum_{k=1}^\infty(-1)^k\frac{\log(2k)}{2k}}^{\substack{+2\leftrightarrow0\pmod4\\-2\leftrightarrow2\pmod4}} &=\sum_{k=1}^\infty(-1)^{k-1}\frac{\log(2)}k\\ &=\log(2)^2 \end{align} $$