Question

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))+...`?

Answer 1

The partial sum is `=3/4-(2n+3)/(2(n+1)(n+2))`
The series converge to `3/4`

Explanation:

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`

Answer 2

`sum_(n=1)^oo 1/(n(n+2)) =3/4`

Explanation:

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`