The sum to infinity of the series (1/3) + (1/(15)) + (1/(35)) + .... is (1/2).


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Question Number 56423 by gunawan last updated on 16/Mar/19

$The sum to infinity of the series$ $\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + . . . . is \frac{1}{2} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19
	$T_1 =(1/2)(((3−1)/(3×1)))=(1/2)(1−(1/3)) T_2 =(1/2)(((5−3)/(5×3)))=(1/2)((1/3)−(1/5)) ... ... T_n =(1/2){(((2n+1)−(2n−1))/((1+(n−1)×2)(3+(n−1)×2)))} =(1/2){(1/(2n−1))−(1/(2n+1))} S_n =(1/2)(1−(1/(2n+1))) S_∞ =(1/2)(1−0)=(1/2)$
	$T_{1} = \frac{1}{2} (\frac{3 - 1}{3 \times 1}) = \frac{1}{2} (1 - \frac{1}{3})$ $T_{2} = \frac{1}{2} (\frac{5 - 3}{5 \times 3}) = \frac{1}{2} (\frac{1}{3} - \frac{1}{5})$ $. . .$ $. . .$ $T_{n} = \frac{1}{2} {\frac{(2 n + 1) - (2 n - 1)}{(1 + (n - 1) \times 2) (3 + (n - 1) \times 2)}}$ $= \frac{1}{2} {\frac{1}{2 n - 1} - \frac{1}{2 n + 1}}$ $S_{n} = \frac{1}{2} (1 - \frac{1}{2 n + 1})$ $S_{\infty} = \frac{1}{2} (1 - 0) = \frac{1}{2}$
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