How do you find a formula for the nth partial sum of the series [5/1*2]+[5/2*3]+[5/3*4]+...+[5/n(n+1)]+... and use it to find the series' sum if the series converges?

1 Answer
Feb 28, 2015

First of all we can factor the #5#:

#sum_1^(+oo)5/(n(n+1))=5sum_1^(+oo)1/(n(n+1))=(1)#

This series, except for the factos #5#, is named Mengoli Series.

Now let's try to separate the term #1/(n(n+1)# in the sum of two terms:

#1/(n(n+1))=A/n+B/(n+1)=(A(n+1)+Bn)/(n(n+1))=#

#=(n(A+B)+B)/(n(n+1)#.

For the identity principle of polynomials:

#1=n(A+B)+B#

So:

#A+B=0#
#A=1#

That gives:

#A=1#
#B=-1#.

So:

#(1)=5sum_1^(+oo)(1/n-1/(n+1))#.

Let's see now #S_n#:

#S_n=5(1/1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1))#,

We can notice now that all the terms except the first and the last will erase themselves, so:

#S_n=5(1-1/(n+1))# and we know that we can obtain the sum of ALL the terms of the series with this limit:

#lim_(nrarr+oo)S_n=lim_(nrarr+oo)5(1-1/(n+1))=5(1-0)=5#.