Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 115736 by gab last updated on 28/Sep/20

Answered by Dwaipayan Shikari last updated on 28/Sep/20

Σ_(n=1) ^∞ (((−1)^(n+1) )/(2n^2 (2i))).(e^(iπn) −e^(−iπn) ) (sin(πn)=((e^(iπn) −e^(−iπn) )/(2i))) (1/(4i))Σ_(n=1) ^∞ (((−1)^(n+1) )/n^2 )e^(iπn) −(1/(4i))Σ^∞ (((−1)^(n+1) )/n^2 ).e^(−iπn) (1/(4i))(−(1/1^2 )−(1/2^2 )−...)−(1/(4i))(−(1/1^2 )−(1/2^2 )−(1/3^2 )−...) =0 Or Another method sin(kπ)=0 (k∈Z) So every term is 0 Sum is zero

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{2n}^{\mathrm{2}} \left(\mathrm{2i}\right)}.\left(\mathrm{e}^{\mathrm{i}\pi\mathrm{n}} −\mathrm{e}^{−\mathrm{i}\pi\mathrm{n}} \right)\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{sin}\left(\pi\mathrm{n}\right)=\frac{\mathrm{e}^{\mathrm{i}\pi\mathrm{n}} −\mathrm{e}^{−\mathrm{i}\pi\mathrm{n}} }{\mathrm{2i}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4i}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{e}^{\mathrm{i}\pi\mathrm{n}} −\frac{\mathrm{1}}{\mathrm{4i}}\overset{\infty} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}^{\mathrm{2}} }.\mathrm{e}^{−\mathrm{i}\pi\mathrm{n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4i}}\left(−\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−...\right)−\frac{\mathrm{1}}{\mathrm{4i}}\left(−\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−...\right) \\ $$$$=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Or} \\ $$$$\mathrm{Another}\:\mathrm{method}\: \\ $$$$\mathrm{sin}\left(\mathrm{k}\pi\right)=\mathrm{0}\:\:\:\:\:\:\left(\mathrm{k}\in\mathbb{Z}\right) \\ $$$$\mathrm{So}\:\mathrm{every}\:\mathrm{term}\:\mathrm{is}\:\mathrm{0} \\ $$$$\mathrm{Sum}\:\mathrm{is}\:\mathrm{zero} \\ $$

Commented by gab last updated on 28/Sep/20

thank you sir.

$$ \\ $$$${thank}\:{you}\:{sir}. \\ $$

Answered by mathmax by abdo last updated on 28/Sep/20

sin(nπ)=0 ⇒Σ_(n=1) ^∞ (((−1)^(n+1) sin(nπ))/(2n^2 ))=0

$$\mathrm{sin}\left(\mathrm{n}\pi\right)=\mathrm{0}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{sin}\left(\mathrm{n}\pi\right)}{\mathrm{2n}^{\mathrm{2}} }=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com