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 #