Question

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?

Answer 1

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`.