Question

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

[infinity]
Σ 8/n^2-1
n=3

Answer 1

The sum converges at:
`(10)/(3)`

Explanation:

Given

`\sum\limits^(\infty)_(n =2) (8)/(n^2 - 1)`

Express the denominator as difference of two squares

`\sum\limits^(\infty)_(n =2) (8)/((n - 1)(n+1))`

Express 8 as 4 * 2

`\sum\limits^(\infty)_(n =2) (4 * 2)/((n - 1)(n+1))`

Rewrite as:

`4 * \sum\limits^(\infty)_(n =2) (2)/((n - 1)(n+1))`

Express 2 as 1 + 1 + 0

`4 * \sum\limits^(\infty)_(n =2) (1+1+0)/((n - 1)(n+1))`

Express 0 as n - n

`4 * \sum\limits^(\infty)_(n =2) (1+1+n - n)/((n - 1)(n+1))`

Rewrite as:

`4 * \sum\limits^(\infty)_(n =2) ((n + 1)-(n - 1))/((n - 1)(n+1))`

Split

`4 * \sum\limits^(\infty)_(n =2) ((n + 1))/((n - 1)(n+1))-((n - 1))/((n - 1)(n+1))`

Cancel out like terms

`4 * \sum\limits^(\infty)_(n =2) (1)/((n - 1))-(1)/((n+1))`

In the above statement, we have:

`a_3 + a_5 = 4[((1)/(2) - (1)/(4)) + ((1)/(4) - (1)/(6))]`

`a_3 + a_5 = 4[((1)/(2) - (1)/(6))]`

Add
`a_7`

`a_3 + a_5 + a_7= 4[((1)/(2) - (1)/(6)) + ((1)/(7 - 1) - (1)/(7+1))]`

`a_3 + a_5 + a_7= 4[((1)/(2) - (1)/(6)) + ((1)/(6) - (1)/(8))]`

`a_3 + a_5 + a_7= 4[((1)/(2) - (1)/(8))]`

Notice that the pattern follows:

`a_3 + a_5 + a_7 + ...... + a_(k)= 4[((1)/(2) - (1)/(k+1))]`

The above represent the odd sums (say S1)

For the even sums, we have:

`4 * \sum\limits^(\infty)_(n =2) (1)/((n - 1))-(1)/((n+1))`

In the above statement, we have:

`a_4 + a_6 = 4[((1)/(3) - (1)/(5)) + ((1)/(5) - (1)/(7))]`

`a_4 + a_6 = 4[((1)/(3) - (1)/(7))]`

Add
`a_8` to both sides

`a_4 + a_6 +a_8 = 4[((1)/(3) - (1)/(7)) + (1)/(7) - (1)/(9)]`

`a_4 + a_6 +a_8 = 4[(1)/(3) - (1)/(9)]`

Notice that the pattern follows:

`a_4 + a_6 + a_8 + ...... + a_(k)= 4[((1)/(3) - (1)/(k+1))]`

The above represent the even sums (say S2)

The total sum (S) is:

`S = S_1 + S_2`

`S =4[((1)/(2) - (1)/(k+1))] + 4[((1)/(3) - (1)/(k+1))]`

Remove all k terms

`S =4[((1)/(2)] + 4[((1)/(3)]`

Open bracket

`S =(4)/(2) + (4)/(3)`

`S =(12 + 8)/(6)`

`S =(20)/(6)`

`S =(10)/(3)`

The sum converges at:
`(10)/(3)`