How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1/(1*3)+1/(2*4)+1/(3*5)+...+1/(n(n+2))+...#?

Let's perform a decomposition into partial fractions

#1/(n(n+2))=A/n+B/(n+2)#

#=(A(n+2)+Bn)/(n(n+2))#

We compare the numerators

#1=A(n+2)+Bn#

When #n=0#, #=>#, #1=2A#, #=>#, #A=1/2#

When #n=-2#, #=>#, #1=-2B#, #=>#, #B=-1/2#

Therefore,

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

#n=1#, #=>#, #1/(1*3)=1/2-1/6#

#n=2#, #=>#, #1/(2*4)=1/4-1/8#

#n=3#, #=>#, #1/(3*5)=1/6-1/10#

#n=4#, #=>#, #1/(4*6)=1/8-1/12#

#n=5#, #=>#, #1/(5*7)=1/10-1/14#

#n=n-2#, #=>#, #1/((n-2)(n))=1/(2(n-2))-1/(2n)#

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

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

Partial sum

#S_n=1/2+1/4-1/(2(n+1))-1/(2(n+2))#

#=3/4-(2n+3)/(2(n+1)(n+2))#

#lim_(n->+oo)S_n=3/4#

The series converge to #3/4#

We can determine that the series is convergent by direct comparison, since for #n >=1#:

#0 < 1/(n(n+2)) < 1/n^2#

and:

#sum_(n=1)^oo 1/n^2 = pi^2/6 #

is convergent.

To determine the sum we can write the general term of the series as:

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

so the the partial sums are:

#s_N = sum_(n=1)^N 1/(n(n+2)) = 1/2sum_(n=1)^N (1/n -1/(n+2)) = 1/2( 1- cancel(1/3) + 1/2 -1/4 +cancel(1/3) -1/5+...+1/(N-2) -cancel(1/N) +1/(N-1) -1/(N+1) +cancel(1/N) -1/(N+2))#

and we can see that in every partial sum, all the intermediate terms cancel two by two except for:

#s_N = 1/2 (1 +1/2 -1/(N+1) -1/(N+2)) = 3/4 -1/2(1/(N+1) + 1/(N+2))#

so that the sum of the series is:

#sum_(n=1)^oo 1/(n(n+2)) = lim_(N->oo) s_N = lim_(N->oo) [3/4 -1/2(1/(N+1) + 1/(N+2))] = 3/4#