# 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)!))+...?

Feb 1, 2017

The ${n}^{\text{th}}$ 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 $n \to \infty$.

#### Explanation:

First, to find the ${n}^{\text{th}}$ 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 ${n}^{\text{th}}$ partial sum, we can now show that the series converges to $1$ as $n \to \infty$.

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)!))

$= 1 - \frac{1}{\infty}$

$= 1 - 0$

$= 1$

Feb 1, 2017

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)!))
$\text{ } = \left(1 - \frac{1}{2}\right) +$
$\text{ } \left(\frac{1}{2} - \frac{1}{6}\right) +$
$\text{ } \left(\frac{1}{6} - \frac{1}{24}\right) + \ldots +$
 " "(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)!))
$\text{ } = 1 - 0$
$\text{ } = 1$