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?
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?
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 $\arctan 2$.
The given series is just the argument of $$ \prod_{n\geq 1}\left(1+\frac{2i}{n^2+n+4}\right) =\left(\frac{4}{5}+\frac{8 i}{5}\right) \text{sinh}(2\pi)\, \text{sech}\left(\frac{\sqrt{15} \pi }{2}\right)$$ i.e. $\color{red}{\arctan 2}$. The above identity follows from the Weierstrass products of the (hyperbolic) sine and cosine functions.
"Recall" this trigonometric identity: $$ \arctan x + \arctan y = \arctan \frac{x+y}{1-xy} $$ Now think about applying that: \begin{align} & \arctan \frac{an+b}{cn+d} + \arctan \frac{en+f}{gn+h} \\[10pt] = {} & \arctan \frac{\dfrac{an+b}{cn+d} + \dfrac{en+f}{gn+h}}{1 - \dfrac{an+b}{cn+d} \cdot \dfrac{en+f}{gn+h}} \\[10pt] = {} & \arctan \frac{(an+b)(gn+h) + (en+f)(cn+d)}{(cn+d)(gn+h) - (an+b)(en+f)} \\[10pt] = {} & \arctan \frac{(ag+eh)n^2 + (bg+ah+ed+fe)n + (bh + fd)}{(cg-ae)n^2 + (ch+dg-af-be)n + (dh-bf)} \end{align} You'd want to choose $a,b,c,d,e,f,g,h$ in such a way that $ag+eh=0,$ $cg-ae=1,$ etc.
Then you've got two terms, $\dfrac{an+b}{cn+d} + \arctan \dfrac{en+f}{gn+h},$ where originally you had one.
The $k$th partial sum is \begin{align*} \sum_{n = 1}^k (\tan^{-1} \sqrt{n} - \tan^{-1} \sqrt{n + 1}) &= (\tan^{-1} 1 - \tan^{-1} \sqrt{2}) + (\tan^{-1} \sqrt{2} - \tan^{-1} \sqrt{3}) \\ &\quad + (\tan^{-1} \sqrt{3} - \tan^{-1} \sqrt{4}) + \dots + (\tan^{-1} \sqrt{k} - \tan^{-1} \sqrt{k + 1}) \\ &= \tan^{-1} 1 - \tan^{-1} \sqrt{k + 1} \\ &= \frac{\pi}{4} - \tan^{-1} \sqrt{k + 1}. \end{align*}As $k$ goes to infinity, $\tan^{-1} \sqrt{k + 1}$ approaches $\frac{\pi}{2},$ so the limit of the sum as $n$ goes to infinity is $\frac{\pi}{4} - \frac{\pi}{2} = \boxed{-\frac{\pi}{4}}.$