How do we calculate this?
$ \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{2n}}{2n} $
I am stuck that the integrals isn't converging for harmonic (even) numbers . somebody please help .
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1Independently of what the solution is, note that the Harmonic numbers grow like $\log(n)$ and thus, as the sum is alternating, the series will converge, by the Leibniz test. – b00n heT Jul 23 '16 at 19:01
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Yup I am also pretty convinced about the convergence , WA also supports but when I find the generating function of even harmonics and then integrate we fond a non-converging integral. – Aditya Narayan Sharma Jul 23 '16 at 19:03
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5This question has different answers here:http://math.stackexchange.com/questions/1000140/evaluating-sum-n-1-infty-1n-1-frach-2nn – Olivier Oloa Jul 23 '16 at 19:06
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1Oh thanks @OlivierOloa :) – Aditya Narayan Sharma Jul 23 '16 at 19:09
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,\,{H_{2n} \over 2n}} & = {H_{2} \over 2} - {H_{4} \over 4} + {H_{6} \over 6} - {H_{8} \over 8} + \cdots = \sum_{n = 0}^{\infty}{H_{4n + 2} \over 4n + 2} - \sum_{n = 0}^{\infty}{H_{4n + 4} \over 4n + 4} \\[5mm] & = \sum_{n = 0}^{\infty}{H_{n + 2} \over n + 2} \,{1 + \pars{-1}^{n} + i^{n} + \pars{-\ic}^{n} \over 4} - \sum_{n = 0}^{\infty}{H_{n + 4} \over n + 4} \,{1 + \pars{-1}^{n} + i^{n} + \pars{-\ic}^{n} \over 4} = \\[5mm] & = \sum_{n = 2}^{\infty}{H_{n} \over n} \,{1 + \pars{-1}^{n} - i^{n} - \pars{-\ic}^{n} \over 4} - \sum_{n = 4}^{\infty}{H_{n} \over n} \,{1 + \pars{-1}^{n} + i^{n} + \pars{-\ic}^{n} \over 4} \\[5mm] & = -\,\half\sum_{n = 1}^{\infty}{H_{n} \over n} \bracks{i^{n} + \pars{-\ic}^{n}} = -\Re\sum_{n = 1}^{\infty}H_{n}\,\ic^{n}\int_{0}^{1}x^{n - 1}\,\dd x \\[5mm] & = -\Re\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n}\,\pars{\ic x}^{n}\,{\dd x \over x} = \Re\int_{0}^{1}{\ln\pars{1 - \ic x} \over x\pars{1 - \ic x}}\,\dd x \\[5mm] & = \color{#f00}{{5 \over 96}\,\pi^{2} - {1 \over 8}\,\ln^{2}\pars{2}} \approx 0.4540 \end{align}
Felix Marin
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