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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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