Math for Non-Geeks/Exercises: Series

{{#invoke:Math for Non-Geeks/Seite|oben}}

Harmonic series

Remark

Analogously, for the generalizes harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$ with $\alpha >1$ we can show:

• $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{(2k-1{)}^{\alpha }}=(1-\frac{1}{2^{\alpha }})\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$
• $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k^{\alpha }}=(1-\frac{1}{2^{\alpha -1}})\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$

Math for Non-Geeks: Template:Aufgabe

Rearrangement theorem for series

{{#invoke:Math for Non-Geeks/Seite|unten}}