Summing the zeta function

Calculus Level 5

Suppose a k = ζ ( 2 k ) , a_k=\zeta (2k), where ζ \zeta is the Riemann zeta function. What is the value of the sum ∑ k = 1 ∞ ζ ( 2 k ) − 1 k \sum \limits_{k=1}^{\infty }\frac{\zeta(2k)-1}{k} ?

Details and assumptions

The Riemann zeta function ζ ( s ) \zeta(s) is defined as ∑ n = 1 ∞ n − s . \sum \limits_{n=1}^{\infty} n^{-s}.


The answer is 0.693147.

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Calvin Lin by Calvin Lin

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2 solutions

Pebrudal Zanu
Jan 4, 2014

First step: ∑ k = 1 ∞ ζ ( 2 k ) − 1 k = ∑ i = 2 ∞ ∑ j = 1 ∞ 1 j × i 2 j \displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k}=\sum_{i=2}^{\infty} \sum_{j=1}^{\infty} \frac{1}{j \times i^{2j}}

Not that:

∫ 0 c 1 1 − x d x = − ln ⁡ ( 1 − c ) = ∑ i = 1 ∞ c i \displaystyle \int_{0}^{c} \frac{1}{1-x} dx=-\ln(1-c)=\sum_{i=1}^{\infty} \frac{c}{i}

If replace c = 1 j 2 c=\frac{1}{j^2}

∑ i = 2 ∞ 1 j 2 × i = − ln ⁡ ( 1 − 1 j 2 ) \displaystyle \sum_{i=2}^{\infty} \frac{1}{j^2\times i}=-\ln \left(1-\frac{1}{j^2}\right)

∑ k = 1 ∞ ζ ( 2 k ) − 1 k = − ∑ n = 2 ∞ ln ⁡ ( 1 − 1 n 2 ) = − ln ⁡ ( ∏ n = 2 ∞ ( 1 − 1 n 2 ) ) \displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k} = -\sum_{n=2}^{\infty} \ln \left(1-\frac{1}{n^2}\right)=-\ln\left(\prod_{n=2}^{\infty} \left(1-\frac{1}{n^2}\right)\right)

∏ n = 2 ∞ ( 1 − 1 n 2 ) = ∏ n = 2 ∞ ( 1 − 1 n ) × ∏ n = 2 ∞ ( 1 + 1 n ) = 1 2 \displaystyle \prod_{n=2}^{\infty} \left(1-\frac{1}{n^2}\right)=\prod_{n=2}^{\infty}\left(1-\frac{1}{n}\right) \times \prod_{n=2}^{\infty}\left(1+\frac{1}{n}\right)=\frac{1}{2}

So, ∑ k = 1 ∞ ζ ( 2 k ) − 1 k = ln ⁡ ( 2 ) \displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k}=\ln(2)

The answer is 0.693 \fbox{0.693}

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Typo for line 5 ∑ i = 1 ∞ 1 j 2 i × i \displaystyle\sum_{i=1}^{\infty} \frac{1}{j^{2i} \times i} :D

pebrudal zanu - 7 years, 5 months ago

i could not get your 2nd and 3rd step....would u plz explain

Soyam Mohanty - 7 years, 4 months ago

Penjelasan yang sangat bagus Zanu! :)

Tunk-Fey Ariawan - 7 years, 4 months ago

Hi, can you please explain the first step?

Thanks!

Pranav Arora - 7 years, 3 months ago

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Write Riemann-zeta function ζ ( 2 k ) \zeta(2k) as ζ ( 2 k ) = ∑ n = 1 ∞ 1 n 2 k = 1 1 2 k + 1 2 2 k + 1 3 2 k + ⋯ = 1 + 1 2 2 k + 1 3 2 k + ⋯ \begin{aligned} \zeta(2k)&=\sum_{n=1}^\infty\frac{1}{n^{2k}}\\ &=\frac{1}{1^{2k}}+\frac{1}{2^{2k}}+\frac{1}{3^{2k}}+\cdots\\ &=1+\frac{1}{2^{2k}}+\frac{1}{3^{2k}}+\cdots \end{aligned} Then, ζ ( 2 k ) − 1 = ∑ n = 1 ∞ 1 n 2 k − 1 = ( 1 + 1 2 2 k + 1 3 2 k + ⋯ ) − 1 = 1 2 2 k + 1 3 2 k + 1 4 2 k + ⋯ = ∑ n = 2 ∞ 1 n 2 k \begin{aligned} \zeta(2k)-1&=\sum_{n=1}^\infty\frac{1}{n^{2k}}-1\\ &=\left(1+\frac{1}{2^{2k}}+\frac{1}{3^{2k}}+\cdots\right)-1\\ &=\frac{1}{2^{2k}}+\frac{1}{3^{2k}}+\frac{1}{4^{2k}}+\cdots\\ &=\sum_{n=2}^\infty\frac{1}{n^{2k}} \end{aligned} Hence, ∑ k = 1 ∞ ζ ( 2 k ) − 1 k = ∑ k = 1 ∞ 1 k ( ζ ( 2 k ) − 1 ) = ∑ k = 1 ∞ 1 k ∑ n = 2 ∞ 1 n 2 k = ∑ k = 1 ∞ ∑ n = 2 ∞ 1 k ⋅ n 2 k \begin{aligned} \sum_{k=1}^\infty\frac{\zeta(2k)-1}{k}&=\sum_{k=1}^\infty\frac{1}{k}(\zeta(2k)-1)\\ &=\sum_{k=1}^\infty\frac{1}{k}\sum_{n=2}^\infty\frac{1}{n^{2k}}\\ &=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac{1}{k\cdot n^{2k}}\\ \end{aligned} Both part are interchangeable. You may also treat sum notation as integral. I hope this help Pranav. :)

Tunk-Fey Ariawan - 7 years, 3 months ago

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Beautifully done Tunk-Fey Ariawan, thanks a lot! :)

But I am not sure about your last statement, what do you mean by "treat sum notation as integral"? Can you please explain some more on this? Thanks!

Continuing with the Pebrudal's solution, he writes:

∑ k = 1 ∞ ζ ( 2 k ) − 1 k = − ∑ n = 2 ∞ ln ⁡ ( 1 − 1 n 2 ) \displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k} = -\sum_{n=2}^{\infty} \ln \left(1-\frac{1}{n^2}\right)

..but

− ∑ n = 2 ∞ ln ⁡ ( 1 − 1 n 2 ) = ∑ n = 2 ∞ ∑ i = 1 ∞ 1 n 2 × i \displaystyle -\sum_{n=2}^{\infty} \ln \left(1-\frac{1}{n^2}\right)=\sum_{n=2}^{\infty} \sum_{i=1}^{\infty} \frac{1}{n^2\times i}

and this is not what Pebrudal showed in his first step.

Pranav Arora - 7 years, 3 months ago

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@Pranav Arora – I think he made a bit of typo. He should write Maclaurin series of natural logarithm like this: − ln ⁡ ( 1 − c ) = ∑ i = 1 ∞ c i i for ∣ c ∣ < 1. -\ln(1-c) = \sum_{i=1}^\infty\frac{c^i}{i}\;\;\;\text{for}\;|c|<1. Thus, if I continue my approach, it turns out to be ∑ k = 1 ∞ ∑ n = 2 ∞ 1 k ⋅ n 2 k = ∑ n = 2 ∞ ∑ k = 1 ∞ ( 1 n 2 ) k k = − ∑ n = 2 ∞ ln ⁡ ( 1 − 1 n 2 ) . \begin{aligned} \sum_{k=1}^\infty\sum_{n=2}^\infty\frac{1}{k\cdot n^{2k}}&=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac{\left(\frac{1}{n^2}\right)^k}{k}\\ &=-\sum_{n=2}^\infty\ln\left(1-\frac{1}{n^2}\right). \end{aligned} For my statement "treat sum notation as integral", I mean the sum notation can be interchanged like integral notation. I think you've already known that. To make clearer, take a look this example: ∫ x = a b ∫ y = c d f ( x , y ) d y d x = ∫ y = c d ∫ x = a b f ( x , y ) d x d y , \int_{x=a}^b\int_{y=c}^d f(x,y)\,dydx=\int_{y=c}^d \int_{x=a}^b f(x,y)\,dxdy, as well as ∑ k = 1 ∞ ∑ n = 2 ∞ 1 k ⋅ n 2 k = ∑ n = 2 ∞ ∑ k = 1 ∞ ( 1 n 2 ) k k . \sum_{k=1}^\infty\sum_{n=2}^\infty\frac{1}{k\cdot n^{2k}}=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac{\left(\frac{1}{n^2}\right)^k}{k}. Sorry for my terrible explanation because I'm not that good on math theory. The way I solve math problems only use intuitive approach.

Tunk-Fey Ariawan - 7 years, 3 months ago

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@Tunk-Fey Ariawan – Excellent explanation Tunk! I should have been careful about the expansion of ln(1-x). :)

This was a very nice problem.

Pranav Arora - 7 years, 3 months ago

is there any proof for line 2? as far as i know the right hand side has no limit..

Tobby Satyarama - 7 years, 5 months ago

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1 1 − x = 1 + x + x 2 + x 3 + . . . \frac{1}{1-x}=1+x+x^2+x^3+... , where ∣ x ∣ < 1 |x|<1

Two side we integration on interval 0 < x < c 0<x<c where this form satisfy if 0 < c < 1 0<c<1 ,

And the result:

− l n ( 1 − c ) = c + 1 2 c + 1 3 c 2 + . . . -ln(1-c)=c+\frac{1}{2} c+\frac{1}{3}c^2+...

Since 0 < 1 j 2 < 1 0<\frac{1}{j^2}<1 , We can change c = 1 j 2 c=\frac{1}{j^2} .

So, right hand side is converges if |c|<1 and have limit.

pebrudal zanu - 7 years, 5 months ago
Tom Van Lier
Jun 8, 2016

I used some identitities to get there.

∑ k = 1 ∞ ζ ( 2 k ) − 1 k = \displaystyle \sum_{k=1}^{\infty} \dfrac{\zeta(2k) - 1}{k} =

2 ∑ k = 1 ∞ ζ ( 2 k ) − 1 2 k \displaystyle 2\sum_{k=1}^{\infty} \dfrac{\zeta(2k) - 1}{2k}

Now

∑ k = 1 ∞ ζ ( 2 k ) − 1 2 k = ∑ k = 2 ∞ ζ ( k ) − 1 k − ∑ k = 2 ∞ ζ ( 2 k − 1 ) − 1 2 k − 1 \displaystyle \sum_{k=1}^{\infty} \dfrac{\zeta(2k) - 1}{2k} = \sum_{k=2}^{\infty} \dfrac{\zeta(k) - 1}{k} - \sum_{k=2}^{\infty} \dfrac{\zeta(2k-1) - 1}{2k - 1} .

This gives :

2 ∑ k = 1 ∞ ζ ( 2 k ) − 1 2 k = \displaystyle 2\sum_{k=1}^{\infty} \dfrac{\zeta(2k) - 1}{2k} =

∑ k = 2 ∞ ζ ( k ) − 1 k + ∑ k = 1 ∞ ζ ( 2 k ) − 1 2 k − ∑ k = 2 ∞ ζ ( 2 k − 1 ) − 1 2 k − 1 = \displaystyle \sum_{k=2}^{\infty} \dfrac{\zeta(k) - 1}{k} + \sum_{k=1}^{\infty} \dfrac{\zeta(2k) - 1}{2k} - \sum_{k=2}^{\infty} \dfrac{\zeta(2k-1) - 1}{2k - 1} =

∑ k = 2 ∞ ζ ( k ) − 1 k + ∑ k = 2 ∞ ( − 1 ) k ζ ( k ) − 1 k = \displaystyle \sum_{k=2}^{\infty} \dfrac{\zeta(k) - 1}{k} + \sum_{k=2}^{\infty} (-1)^k \dfrac{\zeta(k) - 1}{k} =

1 − γ + l n ( 2 ) + γ − 1 = 1 - \gamma + ln(2) + \gamma - 1 = , using two identities from the series related to the Euler–Mascheroni constant .

l n ( 2 ) ln(2) .

Hence :

∑ k = 1 ∞ ζ ( 2 k ) − 1 k = l n ( 2 ) \displaystyle \sum_{k=1}^{\infty} \dfrac{\zeta(2k) - 1}{k} = ln(2) .

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