How would one evaluate the sum from k=0 to k=infinity of ((-1)^(2k-1))/(2k)?
P PaulDirac New member Joined Jan 24, 2010 Messages 5 Jan 24, 2010 #1 How would one evaluate the sum from k=0 to k=infinity of ((-1)^(2k-1))/(2k)?
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Jan 24, 2010 #2 Factor out the 2, first of all. Secondly, all the powers of -1 are odd, and hence equal to -1: Then you have the sum (note k cannot equal zero... error?): 12∑k>1(−1)2k−1k=−12[11+12+13+14+...]\displaystyle \frac{1}{2}\sum_{k>1}\frac{(-1)^{2k-1}}{k} = -\frac{1}{2}\left [ \frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +... \right ]21k>1∑k(−1)2k−1=−21[11+21+31+41+...] Look familiar now?
Factor out the 2, first of all. Secondly, all the powers of -1 are odd, and hence equal to -1: Then you have the sum (note k cannot equal zero... error?): 12∑k>1(−1)2k−1k=−12[11+12+13+14+...]\displaystyle \frac{1}{2}\sum_{k>1}\frac{(-1)^{2k-1}}{k} = -\frac{1}{2}\left [ \frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +... \right ]21k>1∑k(−1)2k−1=−21[11+21+31+41+...] Look familiar now?
