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arctan(n+1) - arctan(n+2) where n varies from 0 to infinity

I'm having trouble figuring out this problem. I've calculated the first 5 terms of the series and ended up with the following:

-.0.32175 - 0.14189 - 0.07677 - 0.07677 - 0.04758 ......

To me, it is apparent that the series is converging on some number (perhaps -1?) however, I'm not sure how to prove this. Any help is appreciated!

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    Yes, sum of the series.2011-01-26

2 Answers 2

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If you are looking for the sum of the series

$ \sum_{n=0}^\infty (\arctan(n+1)-\arctan(n+2)) $

then what you have is a telescoping series. This means that terms will cancel. The first part of the sum is

$ \arctan(1)-\arctan(2)+\arctan(2)-\arctan(3)+\arctan(3)-\arctan(4)+\cdots $

The terms in the middle will cancel, for instance:

$ -\arctan(2)+\arctan(2)=0 $

Thus your $n$th partial sum is

$ S_n=\arctan(1)-\arctan(n+2) $

Taking the limit gives

$ \lim_{n\rightarrow \infty}(\arctan(1)-\arctan(n+2))=\arctan(1)-\frac{\pi}{2}. $

And $\arctan(1)=\frac{\pi}{4}$.

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    That's the best explanation I could have asked for. Unfortunately, my math teacher wasn't able to explain this to me due to the language barrier (silly huh?). Thank you very much2011-01-26
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Of course, what makes this (clearly telescoping) series interesting is that it can be written in not-so-obviously telescoping form; using the formula $\mathrm{arctan}(u)$ + $\mathrm{arctan}(v)$ = $\mathrm{arctan}\bigl({u+v\over 1-uv}\bigr)$, we can write it as $\sum_{n=0}^{\infty}\mathrm{arctan}{1\over (n+1)(n+2)-1}$ — or, shifting indices by one, $\sum_{n=1}^{\infty}\mathrm{arctan}{1\over n^2+n-1}$.