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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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Do you want the sum of the series or the limit of the sequence? – Qiaochu Yuan Jan 26 '11 at 0:40
Do you mean $\sum_{n} (\arctan(n+1) - \arctan(n+2))$ or $\sum_{n} (-1)^{n} \arctan(n+1)$? – t.b. Jan 26 '11 at 0:42
Yes, sum of the series. – Bill M. Jan 26 '11 at 2:18
up vote 3 down vote accepted

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 much – Bill M. Jan 26 '11 at 2:20

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, $\sum_{n=1}^{\infty}\mathrm{arctan}{1\over n^2+n-1}$.

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