Exercises: Series – "Math for Non-Geeks"

Harmonic series

Remark

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

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

Math for Non-Geeks: Template:Aufgabe