Question

What is the sum from of n/2^n? I know by looking at it that it converges absolutely. Actually, I know from wolfram that the sum is 2, I just can't figure out how to show it.

Answer 1

However, unhelpful this may seem, my best guess would be to suppose it equals 2, and prove that it's true. Using epsilon-delta proofs.

Answer 2

fyi, the integral was an absolute beast (for me) and didn't come out to exactly 2, came out to \[\approx\] 2.08 I'm wondering if there's no simple "series" test method to show this.

Answer 3

I'm unfamiliar with epsilon-delta proofs. I'll look it up , thanks

Answer 4

If you're unfamiliar with them, you'll probably have a rather hard time...

Answer 5

An alternative, equally unhelpful tip would be to prove that \[\large \sum_{n=1}^{\infty} {n \over a^n}={a \over(a-1)^2}\]Which seems to be true based on values given to me in wolfram alpha.

Answer 6

Then substitute \(a=2\)

Answer 7

I see the relation, i'm just shocked at the possible methods for solving. Thank you

Answer 8

I've got a way that seems to work using rather basic methods. Give me a minute to write it out.

Answer 9

Use the formula for the sum of a geometric sequence.\[S=a{1-r^x \over 1-r}\]Where r is the common ratio. To find this, just find simplfy\[\Large {{n+1 \over 2^{n+1}} \over {n \over 2^n}}={1+n \over 2n}\]

Answer 10

Then substitute in that expression for r, and 1 for a. And take the limit as n goes to infinity. Then take the limit as x goes to infinity.\[\Large \lim_{x \rightarrow \infty} \left(\lim_{n \rightarrow \infty} {{1-\left({1+n \over 2n}\right)^x} \over {1-{1+n \over 2n}}}\right)\]While a little bit messy, it seems to work.