>>14395858To evaluate this sum we must find a converging sequence of rational lower and upper bounds such that the nth bounds contain the nth partial sum and all partial sums onward.
We can evaluate the partial sums using Algebra:
First, we will call the nth partial sum "S_n":
S_n = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2^n
Next, divide S by 2:
S_n = 1/4 + 1/8 + 1/16 + 1/32 ... + 1/2^(n+1)
Now subtract S/2 from S. All the terms except 1/2 and 1/2^(n+1) cancel out.
And we get:
S_n - S_n/2 = 1/2 - 1/2^(n+1)
Simplify:
S_n/2 = 1/2 - 1/2^(n+1)
And so:
S_n = 1 - 1/2^n
For the lower bounds we can just use the partial sums themselves since they are only getting bigger.
For the upper bounds we can use 1.
To show that the bounds are converging, pick a positive integer M and let's try to find a number n such that our bounds are less than 1/M apart. We can just use n = M. Our nth lower bound is 1 - 1/2^n and our nth upper bound is 1. The difference between the bounds is 1/2^n. Since 2^n > n, 1/2^n < 1/n = 1/M.
Note that all our bounds are consistent with 1. Therefore the number we have just calculated for the sum is equal to 1.
