We are going to calculate pi using the following sum: n=02n+1(1)n=4 Since this summation goes to infinity we can't really evaluate the whole thing, so we'll ask the user how far they want to go, i.e. what L (the upper bound for n's) they want to take the sum to: n=0L2n+1(1)n If you're curious exactly how this works, we start at n=0 and evaluate the formula: (1)0/(2(0)+1)=1/1 (1)1/(2(1)+1)=1/3 (1)2/(2(2)+1)=1/5 (1)3/(2(3)+1)=1/7 and so on until we reach L (inclusively). If we pick L=10 there will be 17 terms, from 0 up to and including 10 . 131+5171+91111+131151+171191+211