Question

Question #88d11

Answer 1

`11/6`

Explanation:

By Partial Fraction Decomposition,

`3/(n(n+3))=1/n-1/(n+3)`

Let us find the partial sum:

`s_n=(1/1-cancel(1/4))+(1/2-cancel(1/5))+(1/3-cancel(1/6))+(cancel(1/4)-cancel(1/7))+cdots+(cancel(1/(n-3))-cancel(1/n))+(cancel(1/(n-2))-1/(n+1))+(cancel(1/(n-1))-1/(n+2))+(cancel(1/n)-1/(n+3))`

By cleaning up the remaining terms,

`=(11)/6-1/(n+1)-1/(n+2)-1/(n+3)`

Now, we can find the sum.

`sum_(n=1)^(infty)3/(n(n+3))=lim_(n to infty)(11/6-1/(n+1)-1/(n+2)-1/(n+3))=11/6-0-0-0=11/6`

I hope that this was clear.