Can You Break This Down?

$\frac{1}{n^{2}+3n+2} = \frac{1}{n+1} - \frac{1}{n+2}$

now for n = 1,2,3,4,.......

$\sum_{n=1}^{\infty} \frac{1}{n+1} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ......$

similarly,

$\sum_{n=1}^{\infty} \frac{1}{n+2} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + ......$

therefore,

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

$\begin{aligned} S &=\lim_{N \rightarrow \infty} \sum_{n=1}^{N}\dfrac{1}{n^2+3n+2}\\ &=\lim_{N \rightarrow \infty} \sum_{n=1}^{N}\dfrac{1}{(n+1)(n+2)}\\ &=\lim_{N \rightarrow \infty} \sum_{n=1}^{N}\dfrac{1}{n+1}-\dfrac{1}{n+2}\\ &=\lim_{N \to \infty} \dfrac{1}{2}-\dfrac{1}{N+2}=\boxed{\dfrac{1}{2}} \end{aligned}$

$\color{#3D99F6}{\sum^{\infty}_{n=1}\frac{1}{n^2+3n+2}=\sum^{\infty}_{n=1}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\\ \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+\dotsm}$ We observe that all the terms after $\color{#D61F06}{\frac{1}{2}}$ cancel out so the answer is $\color{#69047E}{\boxed{\dfrac{1}{2}}}$

We note that:

$\sum _{n=1} ^\infty \dfrac {1}{n^2+3n+2} = \sum _{n=1} ^\infty \dfrac {1}{(n+1)(n+2)} = \sum _{n=1} ^\infty \left( \dfrac {1}{n+1} - \dfrac {1}{n+2} \right)$ $= \left( \frac {1}{2} - \frac {1}{3} \right) + \left( \frac {1}{3} - \frac {1}{4} \right) + \left( \frac {1}{4} - \frac {1}{5} \right) + ... = \boxed {\dfrac {1}{2}}$

A different approach...

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

By the way, I got $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1$ from the Brilliant Telescoping Series material that was linked on the problem description.

Tuccha hai

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

Then, when we add all the terms together, the 2nd and 1st terms will cancel, leaving

$\frac{1}{2}-\frac{1}{\infty+1}=\frac{1}{2}$

very nice solution

Telescopic sum rule

I did It with vn method (nice method read about it people)