find the sum of a series q

**twilightstr** · May 15th 2009, 12:00 AM

find the sum of sigma from n =1 to infinity of [arctan(n+1)- arctann]

**chisigma** · May 15th 2009, 01:16 AM

$\sum_{n=1}^{\infty} \{\tan^{-1} (n+1) - \tan^{-1} (n) \} = \sum_{n=1}^ {\infty} \int_{n}^{n+1} \frac{dx}{1+x^{2}} = \int_{1}^{\infty} \frac{dx}{1+x^{2}} = \frac{\pi}{4}$

Kind regards

$\chi$ $\sigma$

**twilightstr** · May 15th 2009, 01:26 AM

how did u get 1/(1-x)^2?

**chisigma** · May 15th 2009, 01:43 AM

Remember that...

$\int \frac{dx}{1+x^{2}} = \tan^{-1} x + c$

Kind regards

$\chi$ $\sigma$

**Soroban** · May 15th 2009, 02:02 AM

Hello, twilightstr!

Find the sum of: . $S \;=\;\sum^{\infty}_{n=1}\bigg[\arctan(n+1)- \arctan(n)\bigg]$

We have:

. . $S \;=\;\bigg[{\color{red}\rlap{////////}}\arctan(2) - \arctan(1)\bigg] +$ $\bigg[{\color{green}\rlap{////////}}\arctan(3) - {\color{red}\rlap{////////}}\arctan(2)\bigg] +$ $\bigg[{\color{blue}\rlap{////////}}\arctan(4) - {\color{green}\rlap{////////}}\arctan(3)\bigg] + \hdots$

. . $S \;=\;-\arctan(1)$

. . $S\;=\;-\frac{\pi}{4}$

**Spec** · May 15th 2009, 02:29 AM

Aren't you forgetting the term $\arctan \infty$

$S=-\frac{\pi}{4}+\lim_{n \to \infty}\arctan (n+1)=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}$

**Soroban** · May 15th 2009, 06:36 AM

Hello, Spec!

Aren't you forgetting the term $\arctan \infty$ ?

$S\:=\:-\frac{\pi}{4}+\lim_{n \to \infty}\arctan (n+1)=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}$

Yes, I did!

I forgot to look at the other end of the series . . . *blush*

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