Sum $\frac {1}{9} + \frac {1}{18} +\frac {1}{30} +\frac {1}{45} +\frac {1}{63} \ldots$

Question

What is the sum of the following series:

$\frac {1}{9} + \frac {1}{18} +\frac {1}{30} +\frac {1}{45} +\frac {1}{63} \ldots$ I tried to do it through telescope method but it didn't work.

Answer 1

I believe we have,

$$\frac{2}{3} \sum_{n=2}^{\infty} \frac{1}{n(n+1)}$$

Clearly telescoping will work.

How I got it,

Let's look at denominators.

$$9,18,30,45,63,...$$

Take difference of consecutive two.

$$9,12,15,18,..$$

This is clearly given by $3(x-1)+9$ because the difference between every two consecutive numbers is $3$.

So,

$$f(x+1)-f(x)=3x+6$$

Actually shifting everything, the terms, to the right $1$ so that the terms start at $x=2$ gives,

$$f(x+1)-f(x)=3(x-1)+6=3x+3$$

This is convenient to do because it factors,

$$f(x+1)-f(x)=3(x+1)$$

Now sum both sides of the top equation from $x=2$ to $n-1$ and something magical will happen. Keep in mind $f(2)=9$.