Can you solve this operation?

Question

Since I could only put words, check out this link and tell me if you can solve this summation operation. It's the summation of 1/(2^n) where the lower range is n=i and the upper range is infinity.

http://development1.blogspot.com/2007/02/solve-this-operation.html

Answer 1

This is a geometric serie r=1/2
if it starts with 1 its sum would be 2.
As you start with n=i you have to substract the sum of the terms from 1 to i-1

This sum is [1/2^(i-1) *1/2 -1]/(1/2-1)= 2*(1-1/2^i).Substract this from 2 and you get

The result is 2/2^i

Answer 2

Are you sure you don't mean "1" to infinity? Because indexing things from "i" to infinity doesn't make any sense. If it's just 1 to infinity, then that's just the geometeric series (a + ar + ar^2 + ar^3 +...) - 1 where a=1 and r=1/2. In this case the sum is 1/(1-r) - 1, or 1/(1/2) - 1 = 2 - 1 = 1.

If "i" was supposed to be some finite number, in which case you just want to find the sum from that point onwards, then you'd take 1 and subtract the partial sum from 1 to i-1 (or even easier, start with 2 and subtract the sum from 0 to i-1). This other sum would just be another geometric series, albeit a finite one. So the answer would be 2 - (1 - (1/2)^i)/(1 - 1/2), or 2 - 2(1 - (1/2)^i).

I'll go post this answer on the blog too.

Answer 3

I won't even check the link. Here's how to do it.

Rewrite the sum as a constant times the sum from 0 to infinity of the same terms (the constant is just a negative power of 2).

But that new sum just equals 2.

Multiply.

And you're done.

Answer 4

Let S=SUM(1/(2^n)) with n=i to n=infinity

Multiply by 2 and get 2S = SUM(1/(2^n-1)) with n=i to n=infinity

Take 2S - S and you get
S=SUM(1/(2^n-1)) - SUM(1/(2^n)) with n=i to n=infinity
Write out a few terms and you get
S = (1/2^i-1 + 1/2^i + ......+ 0) - (1/2^i + 1/2^i+1 + ........+ 0)
giving S = 1/2^(i-1) = 2/2^i



Ouch, doing maths on a keyboard is hell!