Question

How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given `(1-1/2)+(1/2-1/6)+(1/6-1/24)+...+(1/(n!)-1/((n+1)!))+...`?

Answer 1

The `n^"th"` partial term is given by

`sum_(k=1)^n(1/(k!)-1/((k+1)!)) = 1-1/((n+1)!)`

and the series converges to `1` as `n->oo`.

Explanation:

First, to find the `n^"th"` partial sum:

`sum_(k=1)^n(1/(k!)-1/((k+1)!)) = (1/(1!)-1/(2!))+(1/(2!)-1/(3!))+...+(1/(n!)-1/((n+1)!))`

`=1+(-1/(2!)+1/(2!))+(-1/(3!)+1/(3!))+...+(-1/(n!)+1/(n!))-1/((n+1)!)`

`=1-1/((n+1)!)`

A series like this, in which all but a finite number of terms cancel out, is called a telescoping series.

With our closed form for the `n^"th"` partial sum, we can now show that the series converges to `1` as `n->oo`.

`sum_(k=0)^oo(1/(k!)-1/((k+1)!)) = lim_(n->oo)sum_(k=0)^n(1/(k!)-1/((k+1)!))`

`=lim_(n->oo)(1-1/((n+1)!))`

`=1-1/oo`

`=1-0`

`=1`

Answer 2

The series converges to unity.

Explanation:

The nth partial sum, `S_n`, is given by:

`S_n = sum_(r=1)^(n) (1/(r!)-1/((r+1)!))`
`" "= (1-1/2) +`
`" " (1/2-1/6)+`
`" "(1/6-1/24)+...+`
`" "(1/((n-1)!)-1/(n!)) +`
`" "(1/(n!)-1/((n+1)!))`

Note that almost all the terms cancel, leaving:

`S_n = 1 -1/((n+1)!)`

As `n rarr oo => (n+1)! rarr oo => 1/((n+1)!) rarr 0`

And so,

`lim_(n rarr oo) S_n = lim_(n rarr oo) (1 -1/((n+1)!))`
`" " = lim_(n rarr oo) (1) - lim_(n rarr oo)(1/((n+1)!))`
`" " = 1 - 0`
`" " = 1`