Infinite summation of an infinite summation?

Calculus Level 3

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

The Riemann zeta function is defined as ζ ( s ) = ∑ r = 1 ∞ 1 r s \zeta(s)=\displaystyle \sum_{r=1}^{\infty} \dfrac{1}{r^s} . Find the value of the above expression.


The answer is 1.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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1 solution

Sharky Kesa
May 8, 2016

Substituting the formula of the Riemann Zeta function into the summation, we have

∑ n = 2 ∞ ( ζ ( n ) − 1 ) = ∑ n = 2 ∞ ( ( ∑ r = 1 ∞ 1 r n ) − 1 ) = ∑ n = 2 ∞ ∑ r = 2 ∞ 1 r n = ∑ r = 2 ∞ ∑ n = 2 ∞ 1 r n = ∑ r = 2 ∞ 1 r 2 1 − 1 r = ∑ r = 2 ∞ 1 r ( r − 1 ) = ∑ r = 2 ∞ 1 r − 1 − 1 r = 1 \begin{aligned} \displaystyle \sum_{n=2}^{\infty} (\zeta(n) - 1) &= \displaystyle \sum_{n=2}^{\infty} \left ( \left (\sum_{r=1}^{\infty} \dfrac {1}{r^n} \right )- 1 \right )\\ &= \displaystyle \sum_{n=2}^{\infty} \sum_{r=2}^{\infty} \dfrac {1}{r^n}\\ &= \displaystyle \sum_{r=2}^{\infty} \displaystyle \sum_{n=2}^{\infty} \dfrac {1}{r^n}\\ &= \displaystyle \sum_{r=2}^{\infty} \dfrac{\frac{1}{r^2}}{1-\frac{1}{r}}\\ &= \displaystyle \sum_{r=2}^{\infty} \dfrac{1}{r(r-1)}\\ &= \displaystyle \sum_{r=2}^{\infty} \dfrac{1}{r-1}-\dfrac{1}{r}\\ &=1 \end{aligned}

Thus, the answer is 1.

13 Helpful 0 Interesting 0 Brilliant 0 Confused

Do prove that rearranging is justified.

Jake Lai - 5 years, 1 month ago

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∑ n = 2 ∞ ∑ r = 2 ∞ 1 r n = ( 1 2 2 + 1 3 2 . . . ) + ( 1 2 3 + 1 3 3 . . . ) + ( 1 2 4 + 1 3 4 . . . ) . . . \displaystyle \sum_{n=2}^\infty \displaystyle \sum_{r=2}^\infty \frac {1}{r^n} = (\frac {1}{2^2} + \frac {1}{3^2} ...)+(\frac {1}{2^3} + \frac {1}{3^3} ...)+(\frac {1}{2^4} + \frac {1}{3^4} ...)...

You can see here that there is every fraction has one combination of r and n. If you rearrange these, you get

( 1 2 2 + 1 2 3 + 1 2 4 + . . . ) + ( 1 3 2 + 1 3 3 + 1 3 4 + . . . ) + . . . (\frac {1}{2^2} + \frac {1}{2^3} + \frac {1}{2^4} + ...) + (\frac {1}{3^2} + \frac {1}{3^3} + \frac {1}{3^4} + ...) + ...

Which is basically ∑ r = 2 ∞ ∑ n = 2 ∞ 1 r n \displaystyle \sum_{r=2}^\infty \displaystyle \sum_{n=2}^\infty \frac {1}{r^n}

Serkan Muhcu - 5 years, 1 month ago

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I mean prove that the series is absolutely convergent. Otherwise, it could lead to funny results like 1 1 − 1 2 + 1 3 − 1 4 + … = 0 \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots = 0 . (See Riemann rearrangment theorem .)

Jake Lai - 5 years, 1 month ago

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@Jake Lai – Doesn't the fact that every term in the summation is positive, and a particular permutation of them is shown to sum to 1, prove that it's absolutely convergent?

ben handley - 5 years, 1 month ago

@Jake Lai – It is not an alternating series is it?

Joel Yip - 5 years ago

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