<s>[INST] Sum $\frac {1}{9} + \frac {1}{18} +\frac {1}{30} +\frac {1}{45} +\frac {1}{63} \ldots$ 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. 
 [/INST] 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 $f(2)=9$.
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