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I tried using partial fractions to compute the sum of the series $$ \sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)} $$

Another technique is to turn this series into a definite integral of 0 to 1. but do not know how to do.

Thanks for any help.

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  • $\begingroup$ A similar question can be found here. $\endgroup$ – Lucian Jul 17 '14 at 6:42
  • $\begingroup$ also similar... $\endgroup$ – draks ... Jul 17 '14 at 8:37
  • $\begingroup$ @Lucian How is it similar? $\endgroup$ – Did Jul 17 '14 at 8:40
  • $\begingroup$ @Did: How is it dissimilar ? $\endgroup$ – Lucian Jul 17 '14 at 9:01
  • $\begingroup$ @Lucian As much as $1/((3n+1)(3n+2)(3n+3))$ is dissimilar from $1/(n(3n+1))$. $\endgroup$ – Did Jul 17 '14 at 9:19
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Note that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n(n+1/3)} = \sum_{n=1}^{\infty}\dfrac{9}{3n(3n+1)} = 9\sum_{n=1}^{\infty}\left(\dfrac{1}{3n}-\dfrac{1}{3n+1}\right)$.

Now, let $f(x) = \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{x^{3n}}{3n}-\dfrac{x^{3n+1}}{3n+1}\right)$. Clearly, $f(0) = 0$ and our sum is $9f(1)$.

Also, termwise differentiation yields $f'(x) = \displaystyle\sum_{n=1}^{\infty}\left(x^{3n-1}-x^{3n}\right) = \dfrac{x^2-x^3}{1-x^3} = \dfrac{x^2}{1+x+x^2}$.

Using partial fractions, we get $f(1) = f(0) + \displaystyle\int_{0}^{1}f'(x)\,dx$ $= \displaystyle\int_{0}^{1}\dfrac{x^2}{1+x+x^2}\,dx = \left[x - \dfrac{1}{2}\ln(1+x+x^2) - \dfrac{1}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right) \right]_0^1$ $= 1 - \dfrac{1}{2}\ln 3 - \dfrac{\pi}{6\sqrt{3}}$

Therefore, our sum is $9f(1) = 9 - \dfrac{9}{2}\ln 3 - \dfrac{\pi\sqrt{3}}{2}$.

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  • $\begingroup$ Nice solution. Thanks. $\endgroup$ – Mathsource Jul 17 '14 at 6:46
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You can start with this identity

$$ \sum_{n=1}^{\infty}\frac{x^{n-2/3}}{n} = -\frac{\ln(1-x)}{x^{2/3}}. $$

Now just integrate both sides w.r.t. $x$ from $0$ to $1$ to get the answer

$$ -\pi \,\sqrt {3}/2+9-9\,\ln \left( 3 \right)/2. $$

Note: We used the identity

$$ \sum_{n=1}^{\infty}\frac{x^{n}}{n}=-\ln(1-x). $$

which is easy to prove.

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Hint

$\frac{1}{(n)(n+\frac{1}{3})}=3(\frac{1}{n}-\frac{1}{n+\frac{1}{3}})$

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  • 1
    $\begingroup$ And then ? Could you elaborate the next steps ? $\endgroup$ – Claude Leibovici Jul 17 '14 at 6:42

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