<s>[INST] Evaluation of $\sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$ Find the following sum
$$S= \sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$$
I am not able to make it telescopic series. Could someone help me with this?
 [/INST] This can be rewritten as
$$\sum_{n=1}^\infty \left(\arctan \frac2n - \arctan \frac2{n+1}\right) = \arctan 2.$$

Of course if we mysteriously find this expression, we can verify it using the formula $\arctan x - \arctan y = \arctan \frac{x-y}{1+xy}$. But how do we find this expression in the first place?
If we guess the form of $x$ and $y$, namely, $\frac1{a n + b}$ and $\frac1{a(n+1)+b}$, then we can figure out when $\frac{x-y}{1+xy} = \frac{2}{n^2+n+4}$. This gives us $$\frac{a}{a^2 n^2+a^2 n+2 a b n+a b+b^2+1} = \frac{2}{n^2+n+4}.$$ To get a $2$ in the numerator at the same time as an $n^2$ in the denominator, we want $a = \frac12$, and then $b=0$ follows by straightforward algebra.
(Guessing the form $\frac1{an+b}$ is not too unreasonable, since we want something that goes to $0$ as $n \to \infty.$ It's not the only possibility, of course.)
Another answer is "find the pattern and check". If you compute the first few partial sums and try to stuff them into one $\arctan$ (by using the corresponding addition formula), then we get
$$\arctan \frac13, \arctan \frac47, \arctan \frac34, \arctan \frac89,\arctan  1, \arctan \frac{12}{11},\arctan  \frac{7}{6}, \arctan \frac{16}{13}, \dots,$$ and that's how long it took me to spot the pattern: this is
$$\arctan \frac26,\arctan  \frac47,\arctan  \frac68,\arctan  \frac89,\arctan  \frac{10}{10},\arctan  \frac{12}{11},\arctan  \frac{14}{12},\arctan \frac{16}{13},$$
so the $n^{\text{th}}$ term is $\arctan \frac{2n}{n+5}$. This doesn't let us find the telescoping form, but now we can prove that this is the correct partial sum by induction, and the limit as $n \to \infty$ is $\arctan 2$.
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