Riemann Series Theorem

Author

Eric W. Weisstein
December 22, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Calculus/RiemannSeriesTheorem.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/RiemannSeriesTheorem.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Example

-(∑_k^∞)((((-1))^k)/k)==log(2)

V5

-Sum[(-1)^k/k,{k,∞}]

V7

-Sum[(-1)^k/k,{k,∞}]

(∑_(k=0)^∞)(-(1/(4 k+2))-(1/(4 k+4))+(1/(2 k+1)))==((log(2))/2)

Table[{(1/(2k+1)),-(1/(4k+2)),-(1/(4k+4))},{k,0,4}]

V5

Sum[((1/(2k+1))-(1/(4k+2))-(1/(4k+4))),{k,0,∞}]//Simplify

Sum[((1/(2k-1))-(1/(4k-2))-(1/(4k))),{k,∞}]//Simplify

Simplify[(1/(2k-1))-(1/(4k-2))-(1/(4k))]//Factor

Sum[(1/(4 k (-1+2 k))),{k,∞}]//FullSimplify

V7

Sum[((1/(2k+1))-(1/(4k+2))-(1/(4k+4))),{k,0,∞}]

Sum[((1/(2k-1))-(1/(4k-2))-(1/(4k))),{k,∞}]

Sum[(1/(4 k (-1+2 k))),{k,∞}]

FullSimplify[%]