calculate-n-4-n-n-2-9-2-

Question Number 67010 by mathmax by abdo last updated on 21/Aug/19

c a l c u l a t e \sum_{n = 4}^{+ \infty} \frac{n}{{(n^{2} - 9)}^{2}}

Commented by mathmax by abdo last updated on 22/Aug/19

l e t S = \sum_{n = 4}^{\infty} \frac{n}{{(n^{2} - 9)}^{2}} \Rightarrow S = \sum_{n = 4}^{\infty} \frac{n}{{(n - 3)}^{2} {(n + 3)}^{2}}

b u t {(n + 3)}^{2} - {(n - 3)}^{2} = n^{2} + 6 n + 9 - n^{2} + 6 n - 9 = 12 n \Rightarrow

S = \frac{1}{12} \sum_{n = 4}^{\infty} \frac{{(n + 3)}^{2} - {(n - 3)}^{2}}{{(n + 3)}^{2} {(n - 3)}^{2}} = \frac{1}{12} \sum_{n = 4}^{\infty} \frac{1}{{(n - 3)}^{2}} - \frac{1}{12} \sum_{n = 4}^{\infty} \frac{1}{{(n + 3)}^{2}}

= \frac{1}{12} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - \frac{1}{12} \sum_{n = 7}^{\infty} \frac{1}{n^{2}}

= \frac{1}{12} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - \frac{1}{12} {\sum_{n = 1}^{\infty} \frac{1}{n^{2}} - 1 - \frac{1}{2^{2}} - \frac{1}{3^{2}} - \frac{1}{4^{2}} - \frac{1}{5^{2}} - \frac{1}{6^{2}}}

= \frac{1}{12} (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36})