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infinite sum

DrunkenOldFool

New member
Feb 6, 2012
20
It looks like you guys love to solve infinite series problems. Here are a few more

\( \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^4}\)

\( \displaystyle \sum_{n=1}^{\infty}\frac{(n+1) \cdot (n+1)!}{(n+5)!}\)
 

sbhatnagar

Active member
Jan 27, 2012
95
\( \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^4}\)
\( \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^4}=\zeta(4)=(-1)^2 \frac{B_4 (2\pi)^{4}}{2 \times 4!}=\frac{\pi^4}{90}\)

where $B_4$ is a Bernoulli Number.
 

Sherlock

Member
Jan 28, 2012
59
\( \displaystyle \sum_{n=1}^{\infty}\frac{(n+1) \cdot (n+1)!}{(n+5)!}\)
$\displaystyle \sum_{ n \ge 1}\frac{(n+1)(n+1)!}{(n+5)!} = \lim_{ k \to \infty}\sum_{1 \le n \le k}\frac{(n+1)(n+1)!}{(n+5)(n+4)(n+3)(n+2)(n+1)!} = \lim_{ k \to \infty}\sum_{1 \le n \le k}\frac{(n+1)}{(n+5)(n+4)(n+3)(n+2)}$

By writing the summand in the partial fractions form and denoting the partial sum by $S_{k}$, we find that


$\begin{aligned} S_{k} = \frac{2}{3}\sum_{0 \le n \le k}\frac{1}{n+5}-\frac{3}{2}\sum_{0 \le n \le k}\frac{1}{n+4}+\sum_{0 \le n \le k}\frac{1}{n+3}-\frac{1}{6}\sum_{0 \le n \le k}\frac{1}{n+2}\end{aligned}$


Mapping $n \mapsto n-1$ (replacing $n$ with $n-1$ basically) for the first and the third sum then we find


$\begin{aligned} S_{k} & = \frac{2}{3}\sum_{1 \le n-1 \le k}\frac{1}{n+4}-\frac{3}{2}\sum_{1 \le n \le k}\frac{1}{n+4}+\sum_{1 \le n-1 \le k}\frac{1}{n+2}-\frac{1}{6}\sum_{1 \le n \le k}\frac{1}{n+2} \\& = \frac{2}{3}\sum_{2 \le n \le k+1}\frac{1}{n+4}-\frac{3}{2}\sum_{1 \le n \le k}\frac{1}{n+4}+\sum_{2 \le n \le k+1}\frac{1}{n+2}-\frac{1}{6}\sum_{1 \le n \le k}\frac{1}{n+2} \\& = \frac{2}{3(k+4)}-\frac{2}{15}+\frac{2}{3} \sum_{1 \le n \le k}\frac{1}{n+4}-\frac{3}{2}\sum_{1 \le n \le k}\frac{1}{n+4}+\frac{1}{k+3}-\frac{1}{3}+\sum_{1 \le n \le k}\frac{1}{n+2}-\frac{1}{6}\sum_{1 \le n \le k}\frac{1}{n+2} \\& = \frac{2}{3(k+4)}+\frac{1}{k+3}-\frac{7}{15}-\frac{5}{6}\sum_{1 \le n \le k}\frac{1}{n+4}+\frac{5}{6}\sum_{1 \le n \le k}\frac{1}{n+2} \\& = \frac{2}{3(k+4)}+\frac{1}{k+3}-\frac{7}{15}-\frac{5}{6}\sum_{1 \le n-2 \le k}\frac{1}{n+2}+\frac{5}{6}\sum_{1 \le n \le k}\frac{1}{n+2} \\& = \frac{2}{3(k+4)}+\frac{1}{k+3}-\frac{7}{15}-\frac{5}{6}\sum_{3 \le n \le k+2}\frac{1}{n+2}+\frac{5}{6}\sum_{1 \le n \le k}\frac{1}{n+2} \\& = \frac{2}{3(k+4)}+\frac{1}{k+3}-\frac{7}{15}+\frac{35}{72}-\frac{5}{6(k+4)}-\frac{5}{6(k+3)}-\frac{5}{6}\sum_{1 \le n \le k}\frac{1}{n+2}+\frac{5}{6}\sum_{1 \le n \le k}\frac{1}{n+2} \\& = \frac{2}{3(k+4)}+\frac{1}{k+3}-\frac{5}{6(k+4)}-\frac{5}{6(k+3)}+\frac{7}{360}\end{aligned}$


Where we of coursed mapped $n \mapsto n-2$ for the sum on the left in the fifth line.


Now it's obvious that $\displaystyle \lim_{ k \to \infty}S_{k} = \frac{7}{360}$ and that's value of your sum.
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
\( \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^4}=\zeta(4)=(-1)^2 \frac{B_4 (2\pi)^{4}}{2 \times 4!}=\frac{\pi^4}{90}\)

where $B_4$ is a Bernoulli Number.
Another way to do this is to use Fourier theory on the function $f(x)=x^2\;\;(x\in[-\pi,\pi])$. Its Fourier coefficients are given by $\hat{f}(0)=\pi^2/3$, $\hat{f}(n) = 2(-1)^n/n^2\;(n\ne0).$ The result then follows by applying Parseval's theorem.
 

Krizalid

Active member
Feb 9, 2012
118
Whenever I see factorials, I take advantage of the Gamma and Beta function, so $\displaystyle\frac{(n+1)(n+1)!}{(n+5)!}=\frac{1}{\Gamma (4)}\cdot \frac{(n+1)\Gamma (n+2)\Gamma (4)}{\Gamma (n+6)}=\frac{n+1}{\Gamma (4)}\cdot \beta (n,4)
,$ so the series becomes $\displaystyle\frac{1}{\Gamma (4)}\sum\limits_{n=1}^{\infty }{(n+1)\int_{0}^{1}{{{t}^{n+1}}{{(1-t)}^{3}}\,dt}}=\frac{1}{\Gamma (4)}\int_{0}^{1}{t{{(1-t)}^{3}}\sum\limits_{n=1}^{\infty }{(n+1){{t}^{n}}}\,dt}$ and then the series equals $\displaystyle\frac{1}{\Gamma (4)}\int_{0}^{1}{t\left( (1-t)-{{(1-t)}^{3}} \right)\,dt}$ which achieves the same value as shown.