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Question Number 56423 by gunawan last updated on 16/Mar/19
The sum to infinity of the series  (1/3) + (1/(15)) + (1/(35)) + .... is (1/2).
$$\mathrm{The}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{15}}\:+\:\frac{\mathrm{1}}{\mathrm{35}}\:+\:….\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}}. \\ $$
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)
$${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}−\mathrm{1}}{\mathrm{3}×\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{5}−\mathrm{3}}{\mathrm{5}×\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$… \\ $$$$… \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\left(\mathrm{1}+\left({n}−\mathrm{1}\right)×\mathrm{2}\right)\left(\mathrm{3}+\left({n}−\mathrm{1}\right)×\mathrm{2}\right)}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right\} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$${S}_{\infty} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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