Sum of reciprocals of fourth powers

Calculus Level 2

Find the exact sum of ∑ n = 1 ∞ 1 n 4 = 1 + 1 2 4 + 1 3 4 + 1 4 4 + ⋯ \sum_{n=1}^\infty \frac{1}{n^4}= 1+\frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots also known as ζ ( 4 ) \zeta(4) , where ζ ( s ) \zeta(s) is the Riemann zeta function .


Hint : Follow Euler's famous solution of the Basel problem .

π 4 36 \dfrac{\pi^{4}}{36} π 4 120 \dfrac{\pi^{4}}{120} π 2 12 \dfrac{\pi^{2}}{12} π 4 90 \dfrac{\pi^{4}}{90}

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Yao Liu by Yao Liu , 35, USA

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1 solution

David Stigant
Apr 11, 2019

As before:

sin ⁡ ( x ) x = 1 − x 2 3 ! + x 4 5 ! − x 6 7 ! + . . . = ( 1 − x 2 π 2 ) ( 1 − x 2 4 π 2 ) ( 1 − x 2 9 π 2 ) . . . \frac{{\sin \left( x \right)}}{x} = 1 - \frac{{{x^2}}}{{3!}} + \frac{{{x^4}}}{{5!}} - \frac{{{x^6}}}{{7!}} + ... = \left( {1 - \frac{{{x^2}}}{{{\pi ^2}}}} \right)\left( {1 - \frac{{{x^2}}}{{4{\pi ^2}}}} \right)\left( {1 - \frac{{{x^2}}}{{9{\pi ^2}}}} \right)...

Now, substituting ix for x yields:

sin ⁡ ( i x ) i x = 1 − ( i x ) 2 3 ! + ( i x ) 4 5 ! − ( i x ) 6 7 ! + . . . = ( 1 − ( i x ) 2 π 2 ) ( 1 − ( i x ) 2 4 π 2 ) ( 1 − ( i x ) 2 9 π 2 ) . . . \frac{{\sin \left( {ix} \right)}}{{ix}} = 1 - \frac{{{{\left( {ix} \right)}^2}}}{{3!}} + \frac{{{{\left( {ix} \right)}^4}}}{{5!}} - \frac{{{{\left( {ix} \right)}^6}}}{{7!}} + ... = \left( {1 - \frac{{{{\left( {ix} \right)}^2}}}{{{\pi ^2}}}} \right)\left( {1 - \frac{{{{\left( {ix} \right)}^2}}}{{4{\pi ^2}}}} \right)\left( {1 - \frac{{{{\left( {ix} \right)}^2}}}{{9{\pi ^2}}}} \right)...

Or

sin ⁡ ( i x ) i x = 1 + x 2 3 ! + x 4 5 ! + x 6 7 ! + . . . = ( 1 + x 2 π 2 ) ( 1 + x 2 4 π 2 ) ( 1 + x 2 9 π 2 ) . . . \frac{{\sin \left( {ix} \right)}}{{ix}} = 1 + \frac{{{x^2}}}{{3!}} + \frac{{{x^4}}}{{5!}} + \frac{{{x^6}}}{{7!}} + ... = \left( {1 + \frac{{{x^2}}}{{{\pi ^2}}}} \right)\left( {1 + \frac{{{x^2}}}{{4{\pi ^2}}}} \right)\left( {1 + \frac{{{x^2}}}{{9{\pi ^2}}}} \right)...

So multiplying the first equation by the second yields:

1 + ( x 2 3 ! − x 2 3 ! ) + ( x 4 5 ! − x 4 3 ! 2 + x 4 5 ! ) + . . . = ( 1 − x 2 π 2 ) ( 1 + x 2 π 2 ) ( 1 − x 2 4 π 2 ) ( 1 + x 2 4 π 2 ) ( 1 − x 2 9 π 2 ) ( 1 + x 2 9 π 2 ) . . . 1 + \left( {\frac{{{x^2}}}{{3!}} - \frac{{{x^2}}}{{3!}}} \right) + \left( {\frac{{{x^4}}}{{5!}} - \frac{{{x^4}}}{{3{!^2}}} + \frac{{{x^4}}}{{5!}}} \right) + ... = \left( {1 - \frac{{{x^2}}}{{{\pi ^2}}}} \right)\left( {1 + \frac{{{x^2}}}{{{\pi ^2}}}} \right)\left( {1 - \frac{{{x^2}}}{{4{\pi ^2}}}} \right)\left( {1 + \frac{{{x^2}}}{{4{\pi ^2}}}} \right)\left( {1 - \frac{{{x^2}}}{{9{\pi ^2}}}} \right)\left( {1 + \frac{{{x^2}}}{{9{\pi ^2}}}} \right)...

1 − x 4 90 + . . . = ( 1 − x 4 π 4 ) ( 1 − x 4 16 π 4 ) ( 1 − x 4 81 π 4 ) . . . 1 - \frac{{{x^4}}}{{90}} + ... = \left( {1 - \frac{{{x^4}}}{{{\pi ^4}}}} \right)\left( {1 - \frac{{{x^4}}}{{16{\pi ^4}}}} \right)\left( {1 - \frac{{{x^4}}}{{81{\pi ^4}}}} \right)...

1 − x 4 90 + . . . = 1 − x 4 π 4 ( 1 1 4 + 1 2 4 + 1 3 4 + . . . ) + . . . 1 - \frac{{{x^4}}}{{90}} + ... = 1 - \frac{{{x^4}}}{{{\pi ^4}}}\left( {\frac{1}{{{1^4}}} + \frac{1}{{{2^4}}} + \frac{1}{{{3^4}}} + ...} \right) + ...

equating the x 4 x^4 terms:

− x 4 90 = − x 4 π 4 ( 1 1 4 + 1 2 4 + 1 3 4 + . . . ) - \frac{{{x^4}}}{{90}} = - \frac{{{x^4}}}{{{\pi ^4}}}\left( {\frac{1}{{{1^4}}} + \frac{1}{{{2^4}}} + \frac{1}{{{3^4}}} + ...} \right)

So

π 4 90 = 1 1 4 + 1 2 4 + 1 3 4 + . . . \frac{{{\pi ^4}}}{{90}} = \frac{1}{{{1^4}}} + \frac{1}{{{2^4}}} + \frac{1}{{{3^4}}} + ...

12 Helpful 4 Interesting 8 Brilliant 10 Confused

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