Inspired by Calvin Lin

Calculus Level 3

( 1 4 2 + 1 8 2 + 1 1 2 2 + … ) + ( 1 4 4 + 1 8 4 + 1 1 2 4 + … ) + ( 1 4 6 + 1 8 6 + 1 1 2 6 + … ) + … \begin{aligned} &&\left(\frac{1}{4^2}+\frac{1}{8^2}+\frac{1}{12^2}+\ldots\right) \\& + &\left(\frac{1}{4^4}+\frac{1}{8^4}+\frac{1}{12^4}+\ldots\right) \\ & + & \left(\frac{1}{4^6}+\frac{1}{8^6}+\frac{1}{12^6}+\ldots\right) \\ & + & \ldots \end{aligned}

What is the sum of this series?

Note that the bases are divisible by 4, and the exponents are even.

4 − π 8 \frac{4-\pi}{8} 1 9 \frac{1}{9} 1 3 11 \frac{1}{3\sqrt{11}} 3 − e 2 \frac{3-e}{2}

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Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Jayanta Mandi
May 4, 2015

( 1 4 2 + 1 8 2 + 1 1 2 2 + . . . ) + ( 1 4 4 + 1 8 4 + 1 1 2 4 + . . . ) + . . . ( \frac{1}{ 4^{2} } + \frac{1}{ 8^{2} }+\frac{1}{ 12^{2} }+...)+ ( \frac{1}{ 4^{4} } + \frac{1}{ 8^{4} }+\frac{1}{ 12^{4} }+...)+...

= 1 4 2 ( 1 + 1 4 2 + 1 4 4 + . . . ) + 1 8 2 ( 1 + 1 8 2 + 1 8 4 + . . . ) + . . . = \frac{1}{ 4^{2} } (1+ \frac{1}{ 4^{2} } + \frac{1}{ 4^{4} }+...)+\frac{1}{ 8^{2} } (1+ \frac{1}{ 8^{2} } + \frac{1}{ 8^{4} }+...)+...

= 1 4 2 − 1 + 1 8 2 − 1 + . . . = \frac{1}{ 4^{2} -1} +\frac{1}{ 8^{2} -1}+...

= 1 2 ( 1 3 − 1 5 + 1 7 − 1 9 + . . . ) = \frac{1}{2} ( \frac{1}{3}- \frac{1}{5}+ \frac{1}{7}- \frac{1}{9}+...)

= 1 2 ( 1 − tan ⁡ − 1 1 ) = \frac{1}{2} (1- \tan^{-1} 1 )

= 1 2 ( 1 − π 4 ) = \frac{1}{2} (1- \frac{ \pi }{4} )

16 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Fantastic use of the identity π 4 = 1 − 1 3 + 1 5 − 1 7 + … \frac \pi 4 = 1 - \frac13 +\frac15-\frac17 +\ldots

Can you convert this sum of series as a single summation in terms of Riemann zeta function?

clear and elegant solution!

Otto Bretscher - 6 years, 1 month ago

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Thank you.

Jayanta Mandi - 6 years, 1 month ago

[Response to Challenge Master Note]

We use the definition of Riemann zeta function, which is,

ζ ( i ) = ∑ j = 1 ∞ 1 j i ⟹ ζ ( 2 i ) = ∑ j = 1 ∞ 1 j 2 i \zeta(i)=\sum_{j=1}^\infty\frac{1}{j^i}\implies \zeta(2i)=\sum_{j=1}^\infty\frac{1}{j^{2i}}

The sum (call it S S ) can be written (exactly as it is presented in the problem) compactly using summation notation:

S = ∑ i = 1 ∞ ∑ j = 1 ∞ 1 ( 4 j ) 2 i = ∑ i = 1 ∞ ( 1 1 6 i ( ∑ j = 1 ∞ 1 ( j ) 2 i ) ) = ∑ i = 1 ∞ ζ ( 2 i ) 1 6 i S=\sum_{\color{#D61F06}{i=1}}^\infty\sum_{\color{magenta}{j=1}}^\infty\frac{1}{(4j)^{2i}}=\sum_{\color{#D61F06}{i=1}}^\infty\left(\frac{1}{16^i}\left(\sum_{\color{magenta}{j=1}}^\infty\frac{1}{(j)^{2i}}\right)\right)=\sum_{i=1}^\infty\frac{\zeta(2i)}{16^i}

Prasun Biswas - 6 years, 1 month ago

dide the same, i liked this problem :)

Héctor Andrés Parra Vega - 6 years, 1 month ago

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