Arccot series and Arithmetic progressions?

**fardeen_gen** · May 21st 2009, 09:47 AM

If $x_1,x_2,x_3,\mbox{...}$ are in arithmetic progression with common difference $d$ , show that:

$\mbox{arccot}\left(\frac{1 + x_{1}x_{2}}{d}\right) + \mbox{arccot}\left(\frac{1 + x_{2}x_{3}}{d}\right) + \mbox{arccot}\left(\frac{1 + x_{3}x_{4}}{d}\right) + \mbox{...}$ $+\ \mbox{arccot}\left(\frac{1 + x_{n}x_{n + 1}}{d}\right) = \mbox{arccot}\left(\frac{1 + x_{1}x_{n + 1}}{d}\right)$ , where $x_1 > 0$ and $d > 0$

**Media_Man** · May 22nd 2009, 10:52 AM

Letting $x=\{2,4,6,8,10,12\}$ ,

$\sum_{k=1}^5 \arctan(\frac{2}{1+x_kx_{k+1}}) = .381$

but $\arctan(\frac{2}{1+x_1x_{6}}) = .0798$

**Isomorphism** · May 22nd 2009, 11:12 AM

Originally Posted by fardeen_gen

If $x_1,x_2,x_3,\mbox{...}$ are in arithmetic progression with common difference $d$ , show that:

$\mbox{arccot}\left(\frac{1 + x_{1}x_{2}}{d}\right) + \mbox{arccot}\left(\frac{1 + x_{2}x_{3}}{d}\right) + \mbox{arccot}\left(\frac{1 + x_{3}x_{4}}{d}\right) + \mbox{...}$ $+\ \mbox{arccot}\left(\frac{1 + x_{n}x_{n + 1}}{d}\right) = \mbox{arccot}\left(\frac{1 + x_{1}x_{n + 1}}{d}\right)$ , where $x_1 > 0$ and $d > 0$

Since they are in A.P with common difference 'd', $x_2 - x_1 = x_3 - x_2 = x_4 - x_3 = \cdots = d$

$\sum_{k=1}^{k=n}\mbox{arccot}\left(\frac{1 + x_{k}x_{k+1}}{d}\right) =$ $\sum_{k=1}^{k=n}\mbox{arccot}\left(\frac{1 + x_{k}x_{k+1}}{x_{k+1} - x_k}\right)$ $= \sum_{k=1}^{k=n}\mbox{arccot}(x_{k+1}) - \mbox{arccot}(x_{k}) = \mbox{arccot}(x_{n+1}) - \mbox{arccot}(x_1)$ $=\mbox{arccot}\left(\frac{1 + x_{1}x_{n+1}}{nd}\right)$

**Media_Man** · May 22nd 2009, 11:35 AM

Isomorphism:

$\mbox{arccot}(x_{n+1}) - \mbox{arccot}(x_1)$ = $\mbox{arccot}\left(\frac{1 + x_{1}x_{n+1}}{x_{n+1}-x_{1}}\right)$ = $\mbox{arccot}\left(\frac{1 + x_{1}x_{n+1}}{nd}\right)$

This slight correction will make the theorem true.

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