0
$\begingroup$

Compute the following sum :

S=$\sum_{n=1}^{\infty}\frac{1}{6n^5+15n^4+10n^3-n}$

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

$\endgroup$
  • 1
    $\begingroup$ Please provide the partial fractions decomposition that you determined. $\endgroup$ – MachineLearner Apr 27 '19 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 '19 at 12:39
  • $\begingroup$ Your sum should be $$3/7\,\tan \left( 1/6\,\pi\,\sqrt {21} \right) \pi-6/7\,\Psi \left( 1/2 +1/6\,\sqrt {21} \right) -6/7\,\gamma+{\frac {16\,\ln \left( 2 \right) }{7}} $$ $\endgroup$ – Dr. Sonnhard Graubner Apr 27 '19 at 13:12
  • $\begingroup$ @Dr. Sonnhard Grubner thank you . $\endgroup$ – user668815 Apr 27 '19 at 15:45
2
$\begingroup$

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.

Edit

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 $p$ $$\sum_{n=1}^p\frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}=-72 \left(\psi \left(\frac{5}{3}\right)-2 \psi \left(\frac{7}{4}\right)+\psi \left(\frac{11}{6}\right)\right)-\frac{1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$

$\endgroup$
  • $\begingroup$ Thank you .. But then we find div sum $\endgroup$ – user668815 Apr 27 '19 at 15:45
  • $\begingroup$ @Rozeflowers. Sum from $2$ to $p$ and use asymptotics for large values of $p$. $\endgroup$ – Claude Leibovici Apr 27 '19 at 16:02
  • $\begingroup$ Sum from 1 to p then $\lim_{p\to +\infty} S_p$ ? $\endgroup$ – user668815 Apr 27 '19 at 17:11
  • $\begingroup$ For example : how I find $\sum_1^{\infty}(\frac{1}{n}+\frac{1}{n+1})=?$ $\endgroup$ – user668815 Apr 27 '19 at 17:13
  • $\begingroup$ @Rozeflowers. This is the way. Concerning your last comment, I shall answer tomorrow (here, it is 8:15pm) ! Cheers. $\endgroup$ – Claude Leibovici Apr 27 '19 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy