ban1ka1u

2022-07-10

How do you prove
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(2n-1{\right)}^{s}}=\frac{\left({2}^{s}-1\right)\zeta \left(s\right)}{{2}^{s}}$
where $s>1$

postojahob

Expert

Step 1
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(2n-1{\right)}^{s}}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(2n{\right)}^{s}}=\zeta \left(s\right)$
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(2n-1{\right)}^{s}}+\frac{1}{{2}^{s}}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{s}}=\zeta \left(s\right)$
This gives $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(2n-1{\right)}^{s}}+\frac{1}{{2}^{s}}\zeta \left(s\right)=\zeta \left(s\right)$ . Can you finish? [The first equality folows by splitting $\sum \frac{1}{{n}^{s}}$ into even and odd terms].

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