Question

Determine whether the series is convergent or divergent by expressing sn as a telescoping sum.

Answer 1

Answer:

The sum converges at: $\frac{10}{3}$

Step-by-step explanation:

Given

$\sum\limits^{\infty}_{n =2} \frac{8}{n^2 - 1}$

Express the denominator as difference of two squares

$\sum\limits^{\infty}_{n =2} \frac{8}{(n - 1)(n+1)}$

Express 8 as 4 * 2

$\sum\limits^{\infty}_{n =2} \frac{4 * 2}{(n - 1)(n+1)}$

Rewrite as:

$4 * \sum\limits^{\infty}_{n =2} \frac{2}{(n - 1)(n+1)}$

Express 2 as 1 + 1 + 0

$4 * \sum\limits^{\infty}_{n =2} \frac{1+1+0}{(n - 1)(n+1)}$

Express 0 as n - n

$4 * \sum\limits^{\infty}_{n =2} \frac{1+1+n - n}{(n - 1)(n+1)}$

Rewrite as:

$4 * \sum\limits^{\infty}_{n =2} \frac{(n + 1)-(n - 1)}{(n - 1)(n+1)}$

Split

$4 * \sum\limits^{\infty}_{n =2} \frac{(n + 1)}{(n - 1)(n+1)}-\frac{(n - 1)}{(n - 1)(n+1)}$

Cancel out like terms

$4 * \sum\limits^{\infty}_{n =2} \frac{1}{(n - 1)}-\frac{1}{(n+1)}$

In the above statement, we have:

$a_3 + a_5 = 4[(\frac{1}{2} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{6})]$

$a_3 + a_5 = 4[(\frac{1}{2} - \frac{1}{6})]$

Add $a_7$

$a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{6}) + (\frac{1}{7 - 1} - \frac{1}{7+1})]$

$a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{6}) + (\frac{1}{6} - \frac{1}{8})]$

$a_3 + a_5 + a_7= 4[(\frac{1}{2} - \frac{1}{8})]$

Notice that the pattern follows:

$a_3 + a_5 + a_7 + ...... + a_{k}= 4[(\frac{1}{2} - \frac{1}{k+1})]$

The above represent the odd sums (say S1)

For the even sums, we have:

$4 * \sum\limits^{\infty}_{n =2} \frac{1}{(n - 1)}-\frac{1}{(n+1)}$

In the above statement, we have:

$a_4 + a_6 = 4[(\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7})]$

$a_4 + a_6 = 4[(\frac{1}{3} - \frac{1}{7})]$

Add $a_8$ to both sides

$a_4 + a_6 +a_8 = 4[(\frac{1}{3} - \frac{1}{7}) + \frac{1}{7} - \frac{1}{9}]$

$a_4 + a_6 +a_8 = 4[\frac{1}{3} - \frac{1}{9}]$

Notice that the pattern follows:

$a_4 + a_6 + a_8 + ...... + a_{k}= 4[(\frac{1}{3} - \frac{1}{k+1})]$

The above represent the even sums (say S2)

The total sum (S) is:

$S = S_1 + S_2$

$S =4[(\frac{1}{2} - \frac{1}{k+1})] + 4[(\frac{1}{3} - \frac{1}{k+1})]$

Remove all k terms

$S =4[(\frac{1}{2}] + 4[(\frac{1}{3}]$

Open bracket

$S =\frac{4}{2} + \frac{4}{3}$

$S =\frac{12 + 8}{6}$

$S =\frac{20}{6}$

$S =\frac{10}{3}$

The sum converges at: $\frac{10}{3}$