Hello all! I'm stuck on a problem that I figured wouldn't take me as much time as it has.
Verify that the Cauchy Product of:
Riemann Sum from n=0 to inifinity of (-1)^(n+1)/(n+1) is equal to 2 times the riemann sum from n=0 to inifinity of (-1)^(n-1)/(n-1) (1+1/2+1/3+...1/n)

Alright, so, I know I'm supposed to basically square (-1)^(n+1)/(n+1), focusing on the range from k=0 to k=n. In other words I have:
Riemann Sum from k=0 to k=n of ((-1)^(k+1)(-1)^(n-k))/((k+1)(n-k+1)

I expanded it, but cannot find where the 2 in the final product is supposed to come from. Also, I was unsure if I could add the exponents of the -1 [(k+1)+(n-k)], but figured I couldn't. Can anyone point me in the right direction?