The nth term is n/(k+1+k+2+...+k+n) where k=n(n-1)/2; this works out to n/(nk+n(n+1)/2)= 1/(k+(n+1)/2)= 2/(n²+1). So, we need to calculate the sum for n>0 of 2/(n²+1), which is the same as -1+ the sum for n=-∞ to +∞ of 1/(n²+1). The latter sum equals (proof) πcoth(π), so the final result is πcoth(π)-1. |