Answer:
The sum converges at: [tex]\frac{10}{3}[/tex]
Step-by-step explanation:
Given
[tex]\sum\limits^{\infty}_{n =2} \frac{8}{n^2 - 1}[/tex]
Express the denominator as difference of two squares
[tex]\sum\limits^{\infty}_{n =2} \frac{8}{(n - 1)(n+1)}[/tex]
Express 8 as 4 * 2
[tex]\sum\limits^{\infty}_{n =2} \frac{4 * 2}{(n - 1)(n+1)}[/tex]
Rewrite as:
[tex]4 * \sum\limits^{\infty}_{n =2} \frac{2}{(n - 1)(n+1)}[/tex]
Express 2 as 1 + 1 + 0
[tex]4 * \sum\limits^{\infty}_{n =2} \frac{1+1+0}{(n - 1)(n+1)}[/tex]
Express 0 as n - n
[tex]4 * \sum\limits^{\infty}_{n =2} \frac{1+1+n - n}{(n - 1)(n+1)}[/tex]
Rewrite as:
[tex]4 * \sum\limits^{\infty}_{n =2} \frac{(n + 1)-(n - 1)}{(n - 1)(n+1)}[/tex]
Split
[tex]4 * \sum\limits^{\infty}_{n =2} \frac{(n + 1)}{(n - 1)(n+1)}-\frac{(n - 1)}{(n - 1)(n+1)}[/tex]
Cancel out like terms
[tex]4 * \sum\limits^{\infty}_{n =2} \frac{1}{(n - 1)}-\frac{1}{(n+1)}[/tex]
In the above statement, we have:
[tex]a_3 + a_5 = 4[(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{6})][/tex]
[tex]a_3 + a_5 = 4[(\frac{1}{2} - \frac{1}{6})][/tex]
Add [tex]a_7[/tex]
[tex]a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{6}) + (\frac{1}{7 - 1} - \frac{1}{7+1})][/tex]
[tex]a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{6}) + (\frac{1}{6} - \frac{1}{8})][/tex]
[tex]a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{8})][/tex]
Notice that the pattern follows:
[tex]a_3 + a_5 + a_7 + ...... + a_{k}= 4[(\frac{1}{2} - \frac{1}{k+1})][/tex]
The above represent the odd sums (say S1)
For the even sums, we have:
[tex]4 * \sum\limits^{\infty}_{n =2} \frac{1}{(n - 1)}-\frac{1}{(n+1)}[/tex]
In the above statement, we have:
[tex]a_4 + a_6 = 4[(\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7})][/tex]
[tex]a_4 + a_6 = 4[(\frac{1}{3} - \frac{1}{7})][/tex]
Add [tex]a_8[/tex] to both sides
[tex]a_4 + a_6 +a_8 = 4[(\frac{1}{3} - \frac{1}{7}) + \frac{1}{7} - \frac{1}{9}][/tex]
[tex]a_4 + a_6 +a_8 = 4[\frac{1}{3} - \frac{1}{9}][/tex]
Notice that the pattern follows:
[tex]a_4 + a_6 + a_8 + ...... + a_{k}= 4[(\frac{1}{3} - \frac{1}{k+1})][/tex]
The above represent the even sums (say S2)
The total sum (S) is:
[tex]S = S_1 + S_2[/tex]
[tex]S =4[(\frac{1}{2} - \frac{1}{k+1})] + 4[(\frac{1}{3} - \frac{1}{k+1})][/tex]
Remove all k terms
[tex]S =4[(\frac{1}{2}] + 4[(\frac{1}{3}][/tex]
Open bracket
[tex]S =\frac{4}{2} + \frac{4}{3}[/tex]
[tex]S =\frac{12 + 8}{6}[/tex]
[tex]S =\frac{20}{6}[/tex]
[tex]S =\frac{10}{3}[/tex]
The sum converges at: [tex]\frac{10}{3}[/tex]
