According to Wolfram Mathematica it holds $$ \sum_{k=1}^\infty \frac{1}{4k^2 + 2k} = 1-\log(2) . $$
How can one prove this?
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4What is $\frac{1}{2k} - \frac{1}{2k+1}$? – Daniel Fischer Nov 04 '16 at 15:39
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1Possible duplicate of Evaluating the series $\sum\limits_{n=1}^\infty \frac1{4n^2+2n}$ – Martin Sleziak Feb 02 '17 at 05:37
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See that $${1\over 4k^2 + 2k} = {1\over 2k(2k+1)}= {1\over 2k}-{1\over 2k+1}$$
Therefore $$\begin{align} \sum_{k=1}^{N}{1\over 4k^2 + 2k} &= \sum_{k=1}^{N}{{1\over 2k}-{1\over 2k+1}}\\ &=\sum_{k=1}^{N}{{1\over k}-{1\over 2k+1}-{1\over 2k}}\\ &=H_N-H_{2N}+1 \end{align}$$
Where $H$ is the Harmonic Series. Now since $H_N = \ln N + \gamma + o(1)$ where $\gamma$ is the Euler constant, what follows is that :$$\begin{align} \sum_{k=1}^{N}{1\over 4k^2 + 2k} &=1+H_N-H_{2N} \\ &=1+\ln N-\ln2N+o(1)\\ &=1-\ln\frac{2N}N+o(1)\\ &= 1-\ln 2 + o(1) \end{align}$$
Astyx
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$\sum \dfrac{1}{4k^2+2k}=\sum\dfrac{1}{2k(2k+1)}=\sum \left(\dfrac{1}{2k}-\dfrac{1}{2k+1}\right)=\ldots$
Thomas Andrews
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