Computing the value of $\frac{1}{3^2+1} + \frac{1}{4^2+2} + \frac{1}{5^2+3}\ldots=$?

Question

I have tried converting this series into a telescopic sum whose terms could cancel out but haven't succeeded in that effort. How should I proceed further?

Answer 1

The hint:

Use the telescopic sum and $$(n+2)^2+n=(n+4)(n+1).$$

Now, $$\sum_{n=1}^{+\infty}\frac{1}{(n+2)^2+n}=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{13}{36}.$$

Answer 2

Hint: $$\frac{1}{(n+1)(n+4)}=\frac{1}{3} \Bigg[\frac{1}{n+1}-\frac{1}{n+4}\Bigg]$$

By Cancelling terms we have remaining $$ \frac{1}{3} \Bigg[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\Bigg]= \frac{1}{3} \cdot\frac{13}{12}=\frac{13}{36}$$

Answer 3

You can use:

$$\dfrac{1}{n^2+(n-2)}=\dfrac{1}{(n-1)(n+2)}=\dfrac{1}{3}\left(\dfrac{1}{n-1}-\dfrac{1}{n+2}\right)$$

Answer 4

Note that $\frac{1}{n^2+(n-2)}=\frac{1}{(n-1)(n+2)}=\frac{1}{3}(\frac1{n-1}-\frac1{n+2})$

Therefore the expression becomes $$\frac{1}{3}(\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}...)=\frac13(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})=\frac{36}{13}$$ since everything else cancel out.