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By decompising fractions one can show that \begin{align} \sum_{n=1}^{\infty} \frac{1}{n \, (n+1)^{2} \, (n+3)} = \frac{65}{72} - \frac{\zeta(2)}{2}. \end{align} The fraction can also be seen in the form \begin{align} \frac{1}{n \, (n+1)^{2} \, (n+3)} = \frac{(n+2) \, \Gamma(n)}{(n+1) \, \Gamma(n+4)} = \frac{B(n+1, 3)}{2 \, (n+1)} + \frac{B(n, 4)}{3 \, (n+1)} \end{align} which provides \begin{align} S = \frac{1}{2} \, \sum_{n=1}^{\infty} \frac{B(n+1,3)}{n+1} + \frac{1}{3} \, \sum_{n=1}^{\infty} \frac{B(n,4)}{n+1}. \end{align}

The question proposed is: How are the series to be calculated if the Beta function integral is used to evaluate the series?

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  • $\begingroup$ @BarryCipra It was a typo that has been corrected. $\endgroup$ – Leucippus Jul 30 '15 at 4:05
  • $\begingroup$ Perhaps your solution here is relevant math.stackexchange.com/questions/1378465/… $\endgroup$ – jbuddenh Jul 30 '15 at 4:37
  • $\begingroup$ @jbuddenh It is true that the platform has been presented in another solution. In this case the term $\ln(-1)$ appears in my calculation which does not seem to want to go away. $\endgroup$ – Leucippus Jul 30 '15 at 4:43
  • $\begingroup$ Since the series is real, it makes sense to look at the real parts of the integrals. $\endgroup$ – Math-fun Jul 30 '15 at 10:09
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We have $$\frac{1}{2}\sum_{n\geq1}\frac{B\left(n+1,3\right)}{n+1}=\frac{1}{2}\int_{0}^{1}\left(\sum_{n\geq1}\frac{x^{n}}{n+1}\right)\left(1-x\right)^{2}dx=$$ $$=-\frac{1}{2}\int_{0}^{1}\frac{\log\left(1-x\right)}{x}\left(1-x\right)^{2}dx-\frac{1}{2}\int_{0}^{1}\left(1-x\right)^{2}dx= $$ $$=-\frac{1}{2}\int_{0}^{1}\log\left(u\right)\frac{u^{2}}{1-u}du-\frac{1}{6}=$$ $$=-\frac{1}{2}\sum_{k\geq0}\int_{0}^{1}\log\left(u\right)u^{k+2}du-\frac{1}{6}= $$ $$=\frac{\zeta\left(2\right)}{2}-\frac{19}{24}\tag{1}. $$ For the second sum we have $$\frac{1}{3}\sum_{n\geq1}\frac{B\left(n,4\right)}{n+1}=\frac{1}{3}\int_{0}^{1}\left(\sum_{n\geq1}\frac{x^{n-1}}{n+1}\right)\left(1-x\right)^{3}dx=\frac{1}{3}\int_{0}^{1}\left(\frac{-\log\left(1-x\right)-x}{x^{2}}\right)\left(1-x\right)^{3}dx. $$ Integrating by parts we get $$ =\frac{1}{3}+\int_{0}^{1}\frac{\log\left(1-x\right)\left(1-x\right)^{3}}{x}dx $$ and this integral can be calculated in a similar manner than $(1)$. So we have $$\frac{1}{2}\sum_{n\geq1}\frac{B\left(n+1,3\right)}{n+1}+\frac{1}{3}\sum_{n\geq1}\frac{B\left(n,4\right)}{n+1}=\frac{\zeta\left(2\right)}{2}-\frac{19}{24}+\frac{1}{3}+\frac{49}{36}-\zeta\left(2\right)=\frac{65}{72}-\frac{\zeta\left(2\right)}{2}. $$

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    $\begingroup$ From the result demonstrated the error is Wolfram Alfa. If the integral $$\int \frac{(1-x)^{3} \, \ln(1-x)}{x^2} \, dx$$ is entered into WA the result provided leads to terms of the form $\ln(1-x)$ and $\ln(x-1)$ and more. The logarithm $\ln(x-1)$ evaluated at $x=0$ lead to the problem. This is one of the many examples WA presents errors. $\endgroup$ – Leucippus Jul 30 '15 at 20:17
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    $\begingroup$ WA is a useful tool, but only with an "human verification" behind :) $\endgroup$ – Marco Cantarini Jul 30 '15 at 20:21

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