Question

Is the series convergent or divergent?

Determine whether the series is convergent or divergent by expressing Sn as a telescoping sum.
`sum_(n=5)^oo6/(n^2-1)`

If it is convergent, find its sum.

Answer 1

`27/20`

Explanation:

`sum_(n=5)^oo6/(n^2-1)`

First, express `6/(n^2-1)` using partial fractions.
`6/(n^2-1)=A/(n-1)+B/(n+1)`
`6=A(n+1)+B(n-1)`

Substitute `n=1` to get `A=3` and `n=-1` to get `B=-3`.

Thus, `sum_(n=5)^oo6/(n^2-1)=sum_(n=5)^oo3/(n-1)-3/(n+1)`.

Rewrite this using limit notation: `lim_(L->oo)sum_(n=5)^L3/(n-1)-3/(n+1)`.

Find the first few terms: `lim_(L->oo)((3/4-3/6)+(3/5-3/7)+(3/6-3/8)+(3/7-3/9)+\ldots+(3/(L-1)-3/(L+1)))`.

Rearrange the terms: `lim_(L->oo)(3/4+3/5+(-3/6+3/6)+(-3/7+3/7)+(-3/8+3/8)+\ldots+(-3/(L-1)+3/(L-1))+(-3/L+3/L)-3/(L+1)))`.

Most of the terms cancel out, leaving `lim_(L->oo)3/4+3/5-3/(L+1)`.

This equals `27/20`.