Question

I need help with this summation

Answer 1

\[\LARGE \sum_{r=1}^{\infty} \tan^{-1}(1+r^2+r)-\tan^{-1}(1+r^2-r)\]

Answer 2

i don't understand the canellation pattern to be honest

Answer 3

@hartnn

Answer 4

@Zarkon

Answer 5

\[\sum_{r=1}^\infty\left[\tan^{-1}\left(1+r^2+r\right)-\tan^{-1}\left(1+r^2-r\right)\right]\]
First term: \(\color{red}{\tan^{-1}3}-\tan^{-1}1\)
Second term: \(\color{blue}{\tan^{-1}7}-\color{red}{\tan^{-1}3}\)
Third term: \(\color{green}{\tan^{-1}13}-\color{blue}{\tan^{-1}7}\)
and so on.

The cancellation pattern is there, but I'm confused as to why this leaves you with a negative number. A check with WolframAlpha says the answer should be \(\dfrac{\pi}{4}\), or \(\tan^{-1}1\).

Answer 6

that is EXACTLY what confuses me tbh i aws getting -ve too

Answer 7

was*

Answer 8

@ganeshie8

Answer 9

to butcher notation..you get \[\tan^{-1}(\infty)-\tan^{-1}(1)=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\]

Answer 10

butcher notation?

Answer 11

In other words, the \(n\)-th term of the sequence approaches \(\dfrac{\pi}{2}\), i.e. \(\displaystyle\lim_{n\to\infty}\tan^{-1}n=\frac{\pi}{2}\).

So you're left with \(\dfrac{\pi}{2}-\tan^{-1}1=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}\).