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)!))+...#?

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#

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 #