# 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)!))+...#?

##### 2 Answers

The

and the series converges to

#### Explanation:

First, to find the

#=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

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

#=1-1/oo#

#=1-0#

#=1#

The series converges to unity.

#### Explanation:

The nth partial sum,

# 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

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 #