Riemannsum
Moderatorer: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
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mrcreosote
- Guru
- Innlegg: 1995
- Registrert: 10/10-2006 20:58
Finn [tex]\sum_{k=2}^\infty (\zeta(k)-1)[/tex] der [tex]\zeta[/tex] er Riemanns zetafunksjon.
La
[tex]S=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}(\frac{1}{k})^n[/tex]
Antar at vi kan skifte rekkefølgen på summen:
[tex]S=\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}(\frac{1}{k})^n=\sum_{k=2}^{\infty}\frac{1}{1-\frac{1}{k}}-1-\frac{1}{k}=\sum_{k=2}^{\infty}\frac{k^2}{k(k-1)}-\frac{(k+1)(k-1)}{k(k-1)}=\sum_{k=2}^{\infty}\frac{1}{k(k-1)}=\sum_{k=2}^{\infty}\frac{1}{k-1}-\frac{1}{k}=1[/tex]
[tex]S=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}(\frac{1}{k})^n[/tex]
Antar at vi kan skifte rekkefølgen på summen:
[tex]S=\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}(\frac{1}{k})^n=\sum_{k=2}^{\infty}\frac{1}{1-\frac{1}{k}}-1-\frac{1}{k}=\sum_{k=2}^{\infty}\frac{k^2}{k(k-1)}-\frac{(k+1)(k-1)}{k(k-1)}=\sum_{k=2}^{\infty}\frac{1}{k(k-1)}=\sum_{k=2}^{\infty}\frac{1}{k-1}-\frac{1}{k}=1[/tex]