Well, I am trying to solve find a sum of arctan, but can't find a way.
Can somebody give me a hint?
$\Sigma^\infty_{n=1}tan^{-1}(n+1)-tan^{-1}(n)$
I have tried to integrate it but it seemed way too complicated. I hope there is another way.
The answer is $\frac{\pi}{4}$ by the way.
Hint: Use telescopic formula $$\sum_m^n f(k+1)-f(k)=f(n+1)-f(m)$$
Details: With partial sums $$s_n=\sum_{k=1}^n \tan^{-1}(k+1)-\tan^{-1}(k)=\tan^{-1}(n+1)-\dfrac{\pi}{4}$$ then \begin{align} \sum_{k=1}^\infty \tan^{-1}(k+1)-\tan^{-1}(k) &= \lim_{n\to\infty}s_n \\ &= \lim_{n\to\infty}\left(\tan^{-1}(n+1)-\dfrac{\pi}{4}\right) \\ &= \dfrac{\pi}{2}-\dfrac{\pi}{4} \\ &= \color{blue}{\dfrac{\pi}{4}} \end{align}
Just simply Expand your whole series to get: $$ \sum_{n=1}^{\infty} arctan(n+1) - arc tan(n) $$ $$= arctan (2)-arc tan(1)+ arc tan(3)-arctan(2). . . . . arc tan (\infty) $$
Now , notice that every term is getting cancelled with its additive inverse except $arctan(\infty)$ and $-arctan(1) $
Sum them to get your required answer!
