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Find-n-1-1-2n-1-4-




Question Number 1539 by 314159 last updated on 17/Aug/15
Find Σ_(n=1) ^∞ (1/((2n−1)^4 )).
$$\boldsymbol{\mathrm{Find}}\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}\right)^{\mathrm{4}} }. \\ $$
Answered by 123456 last updated on 17/Aug/15
Σ_(n=1) ^(+∞) (1/n^4 )=Σ_(n=1) ^(+∞) (1/((2n−1)^4 ))+Σ_(n=1) ^(+∞) (1/((2n)^4 ))  Σ_(n=1) ^(+∞) (1/n^4 )=Σ_(n=1) ^(+∞) (1/((2n−1)^4 ))+(1/(16))Σ_(n=1) ^(+∞) (1/n^4 )  Σ_(n=1) ^(+∞) (1/((2n−1)^4 ))=Σ_(n=1) ^(+∞) (1/n^4 )−(1/(16))Σ_(n=1) ^(+∞) (1/n^4 )  Σ_(n=1) ^(+∞) (1/((2n−1)^4 ))=((15)/(16))Σ_(n=1) ^(+∞) (1/n^4 )  Σ_(n=1) ^(+∞) (1/((2n−1)^4 ))=((15)/(16))×(π^4 /(90))=((15π^4 )/(16×90))=((5π^4 )/(16×30))=(π^4 /(16×6))=(π^4 /(96))  Σ_(n=1) ^(+∞) (1/((2n−1)^4 ))=(π^4 /(96))
$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{4}} }+\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{4}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{16}}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{4}} }=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{16}}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{4}} }=\frac{\mathrm{15}}{\mathrm{16}}\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{4}} }=\frac{\mathrm{15}}{\mathrm{16}}×\frac{\pi^{\mathrm{4}} }{\mathrm{90}}=\frac{\mathrm{15}\pi^{\mathrm{4}} }{\mathrm{16}×\mathrm{90}}=\frac{\mathrm{5}\pi^{\mathrm{4}} }{\mathrm{16}×\mathrm{30}}=\frac{\pi^{\mathrm{4}} }{\mathrm{16}×\mathrm{6}}=\frac{\pi^{\mathrm{4}} }{\mathrm{96}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{4}} }=\frac{\pi^{\mathrm{4}} }{\mathrm{96}} \\ $$
Commented by 314159 last updated on 17/Aug/15
Thank a lot...
$$\boldsymbol{\mathrm{Thank}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{lot}}… \\ $$
Commented by 112358 last updated on 17/Aug/15
Out of curiosity, how is                   Σ_(n=1) ^(+∞) (1/n^4 )=(π^4 /(90))  ?
$${Out}\:{of}\:{curiosity},\:{how}\:{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }=\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:\:?\: \\ $$$$ \\ $$
Commented by 123456 last updated on 17/Aug/15
rieman zeta function, later i answer you
$$\mathrm{rieman}\:\mathrm{zeta}\:\mathrm{function},\:\mathrm{later}\:\mathrm{i}\:\mathrm{answer}\:\mathrm{you} \\ $$
Commented by 112358 last updated on 17/Aug/15
Thanks
$${Thanks}\: \\ $$

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