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A 'not too challenge' question...

chisigma

Well-known member
Feb 13, 2012
1,704
The challenging aspect to the question is the unexspected semplicity of the final result...

Find the sum of the series...

$\displaystyle \sum_{n=2}^{\infty} \{1-\zeta(n)\}$ (1)

... where...

$\displaystyle \zeta(s)= \sum_{k=1}^{\infty} \frac{1}{k^{s}}$ (2)

... is the Riemann Zeta Function...

Kind regards

$\chi$ $\sigma$
 

melese

Member
Feb 24, 2012
27
The challenging aspect to the question is the unexspected semplicity of the final result...

Find the sum of the series...

$\displaystyle \sum_{n=2}^{\infty} \{1-\zeta(n)\}$ (1)

... where...

$\displaystyle \zeta(s)= \sum_{k=1}^{\infty} \frac{1}{k^{s}}$ (2)

... is the Riemann Zeta Function...

Kind regards

$\chi$ $\sigma$
$\displaystyle\sum_{n=2}^{\infty }(1-\zeta(n))=\sum_{n=2}^{\infty }\sum_{k=2}^{\infty }\frac{1}{k^n}=\sum_{k=2}^{\infty }\sum_{n=2}^{\infty }\frac{1}{k^n}=\sum_{k=2}^{\infty }\frac{1}{k^2}(1+\frac{1}{k}+\frac{1}{k^2}+\cdots)=\sum_{k=2}^{\infty }\frac{1}{k^2}\cdot\frac{k}{k-1}=\sum_{k=2}^{\infty }\frac{1}{k(k-1)}=\sum_{k=1}^{\infty }\frac{1}{k(k+1)}=1$, where the last equality it by telescoping:$\displaystyle\sum_{k=1}^{t}\frac{1}{k(k+1)}=\sum_{k=1}^{t}( \frac{1}{k}-\frac{1}{k+1})=\frac{1}{1}-\frac{1}{t+1}\to1$ as $t\to\infty$.
 

chisigma

Well-known member
Feb 13, 2012
1,704
The real 'challenge' was in the fact that the correct result is...

$\displaystyle \sum_{n=2}^{\infty} \{1-\zeta(n)\}= -\sum_{n=2}^{\infty}\sum_{k=2}^{\infty} \frac{1}{k^{n}}=...=-1$

... that isn't 1 of course (Wasntme)...

Kind regards

$\chi$ $\sigma$