calculus-find-i-n-2-1-n-n-2-1-ii-n-2-1-n-n-4-1-

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Question Number 131286 by mnjuly1970 last updated on 03/Feb/21

$$\dots calculus\dots .$$$$find::i::\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2-1}=?$$$$ii::\underset{n=2}{\sum ^{\mathrm{\infty }}}\left(\frac{{(-1)}^n}{n^4-1}\right)=?$$$$$$$$$$

Answered by Dwaipayan Shikari last updated on 03/Feb/21

$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2-1}=\frac{1}{3}-\frac{1}{8}+\frac{1}{15}-\frac{1}{24}+\frac{1}{35}-\frac{1}{48}+\dots$$$$=(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+..)-(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\dots )$$$$=\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\dots )-\frac{1}{8}(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\dots )$$$$=\frac{1}{2}-\frac{1}{8}(\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2}{n}-\frac{1}{n+1})$$$$=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$

Commented by mnjuly1970 last updated on 03/Feb/21

$$thanksalotmrpayan$$

Answered by Olaf last updated on 03/Feb/21

$${\mathrm{S}}_1=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2-1}$$$${\mathrm{S}}_1=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(n-1)(n+1)}$$$${\mathrm{S}}_1=\frac{1}{2}[\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n-1}-\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n+1}]$$$${\mathrm{S}}_1=\frac{1}{2}[\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{n}-\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n}]$$$${\mathrm{S}}_1=\frac{1}{2}[-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}+\underset{n=3}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}]$$$${\mathrm{S}}_1=\frac{1}{2}[1-\frac{1}{2}]=\frac{1}{4}$$$$$$$${\mathrm{S}}_2=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^4-1}$$$${\mathrm{S}}_2=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(n^2-1)(n^2+1)}$$$${\mathrm{S}}_2=\frac{1}{2}[\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2-1}-\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2+1}]$$$${\mathrm{S}}_2=\frac{1}{2}[{\mathrm{S}}_1-\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2+1}]$$$${\mathrm{S}}_2=\frac{1}{8}-\frac{1}{2}\{-\frac{i}{4}[-\mathrm{\Psi }(1-\frac{i}{2})+\mathrm{\Psi }(1+\frac{i}{2})$$$$+\mathrm{\Psi }(\frac{3}{2}-\frac{i}{2})-\mathrm{\Psi }(\frac{3}{2}+\frac{i}{2})]\}$$$${\mathrm{S}}_2=\frac{1}{8}-\frac{1}{2}\left\{\frac{\pi }{2\sinh \pi }\right\}$$$${\mathrm{S}}_2=\frac{1}{8}-\frac{\pi }{4\sinh \pi }$$$$$$$$$$

Commented by mnjuly1970 last updated on 03/Feb/21

$$merceyandgratefulmrolaf\dots$$

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