Compute 1≤j<i∑2i+j1,where the sum is taken over all positive integers i and j such that 1≤j<i.
Solution — click to reveal
We have that 1≤j<i∑2i+j1=j=1∑∞i=j+1∑∞2i+j1=j=1∑∞2j1i=j+1∑∞2i1=j=1∑∞2j1(2j+11+2j+21+2j+31+⋯)=j=1∑∞2j1⋅1−1/21/2j+1=j=1∑∞2j1⋅2j1=j=1∑∞4j1=1−1/41/4=31.