Compute the following sum :


My attempt : Use partial fraction :

$6n^5+15n^4+10n^3-n=n(n+1)(2n+1)(3n^2+3n-1)$ $S=\sum_{n=1}^{\infty}(\frac{9(2n+1)}{7(3n^2+3n-1)}-\frac{1}{n}-\frac{1}{1+n}+\frac{16}{7(2n+1)})$ Then use identity digamma " sum " But I find divergence sum

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    $\begingroup$ Please provide the partial fractions decomposition that you determined. $\endgroup$ Apr 27, 2019 at 12:33
  • $\begingroup$ If you multiply $n(n+1)(2n+1)(n^3+3n-1)$, the coefficient of $n^6$ is $2$,but original polynomial has $6$ and $n^5$. Try to fix this. $\endgroup$
    – rtybase
    Apr 27, 2019 at 12:39
  • $\begingroup$ Your sum should be $$3/7\,\tan \left( 1/6\,\pi\,\sqrt {21} \right) \pi-6/7\,\Psi \left( 1/2 +1/6\,\sqrt {21} \right) -6/7\,\gamma+{\frac {16\,\ln \left( 2 \right) }{7}} $$ $\endgroup$ Apr 27, 2019 at 13:12
  • $\begingroup$ @Dr. Sonnhard Grubner thank you . $\endgroup$
    – user668815
    Apr 27, 2019 at 15:45

1 Answer 1


Hint $$6n^5+15n^4+10n^3-n=6n(n+1)\left(n+\frac12 \right)\left(n+\frac{\sqrt{21}+3}{6} \right)\left(n-\frac{\sqrt{21}-3}{6} \right)$$ Use partial fraction decomposition and generalized harmonic numbers or polygamma functions.


Whatever the denominator will be, after partial fraction decomposition, you have $$\frac 1 {P_k(n)}=\sum_{i=1}^k \frac {a_i}{n-r_i}$$ and $$S_p=\sum_{n=1}^p \frac {1}{n-r_i}=\psi (p+1-r_i)-\psi (1-r_i)$$ Now, using asymptotics $$S_p=\log (p)-\psi (1-r_i)+\frac{1-2r_i}{2p}-\frac{6 r_i^2-6 r_i+1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ You just need to recombine everything and continue with Taylor expansions to get not only the limit but also how it is approached.

For sure, the limit will be finite only if $\sum_{i=1}^k a_i=0$.

As an example, consider $$\sum_{n=1}^\infty \frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}$$ $$\frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}=\frac{216}{3 n+2}-\frac{576}{4 n+3}+\frac{432}{6 n+5}$$ $$\sum_{n=1}^p\frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}=72 \psi \left(p+\frac{5}{3}\right)-144 \psi \left(p+\frac{7}{4}\right)+72 \psi \left(p+\frac{11}{6}\right)-72 \psi \left(\frac{11}{6}\right)+144 \psi \left(\frac{7}{4}\right)-72 \psi \left(\frac{5}{3}\right)$$ So, using the asymptotics, for infintely large values of $p$ $$\sum_{n=1}^p\frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}=-72 \left(\psi \left(\frac{5}{3}\right)-2 \psi \left(\frac{7}{4}\right)+\psi \left(\frac{11}{6}\right)\right)-\frac{1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$

  • $\begingroup$ Thank you .. But then we find div sum $\endgroup$
    – user668815
    Apr 27, 2019 at 15:45
  • $\begingroup$ @Rozeflowers. Sum from $2$ to $p$ and use asymptotics for large values of $p$. $\endgroup$ Apr 27, 2019 at 16:02
  • $\begingroup$ Sum from 1 to p then $\lim_{p\to +\infty} S_p$ ? $\endgroup$
    – user668815
    Apr 27, 2019 at 17:11
  • $\begingroup$ For example : how I find $\sum_1^{\infty}(\frac{1}{n}+\frac{1}{n+1})=?$ $\endgroup$
    – user668815
    Apr 27, 2019 at 17:13
  • $\begingroup$ @Rozeflowers. This is the way. Concerning your last comment, I shall answer tomorrow (here, it is 8:15pm) ! Cheers. $\endgroup$ Apr 27, 2019 at 18:15

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