How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #3/(1^2*2^2)+5/(2^2*3^2)+...+(2n+1)/(n^2(n+1)^2)+...#?

1 Answer
Narad T.
Feb 4, 2017

The partial sum is #=(n(n+2))/(n+1)^2#
The series converge and the sum is #=1#

Explanation:

Let's perform a partial fraction decomposition

#(2n+1)/(n^2(n+1)^2)=A/n^2+B/n+C/(n+1)^2+D/(n+1)#

#=(A(n+1)^2+Bn(n+1)^2+Cn^2+Cn^2(n+1))/(n^2(n+1)^2)#

As the denominators are the same, we compare the numerators

#2n+1=A(n+1)^2+Bn(n+1)^2+Cn^2+Dn^2(n+1)#

Let #n=0#, #=>#, #1=A#

Let #n=-1#, #=>#, #-1=C#

Coefficients of #n^3#, #=>#, #0=B+D#

Coefficients of #n#, #=>#, #2=2A+B#

#B=2-2A=2-2=0#

#D=-B=0#

So,

#(2n+1)/(n^2(n+1)^2)=1/n^2-1/(n+1)^2#

Now we can calculate the sums

#n=1#,#=>#,#S_1=1-cancel(1/4)#

#n=2#,#=>#,#S_2=cancel(1/4)-cancel(1/9)#

#n=3#, #=>#, #S_3=cancel(1/9)-cancel(1/16)#
.
.
.

#n=n-1#, #=>#, #S_(n-1)=cancel(1/(n-1)^2)-cancel(1/n^2)#

#n=n#, #=>#, #S_n=cancel(1/n^2)-1/(n+1)^2#

So,

#S_1+S-2+S_2...........S_n=1-1/(n+1)^2#

#=((n+1)^2-1)/(n+1)^2#

#=(n(n+2))/(n+1)^2#

#lim_(n->+oo)(n(n+2))/(n+1)^2#

#=lim_(n->+oo)n^2/n^2=1#

Related questions
See all questions in Partial Sums of Infinite Series
Impact of this question
1942 views around the world
You can reuse this answer
Creative Commons License