Question

Sumfunction

Answer 1

Can someone help me find the sum function to this series
\[\sum_{n=0}^{∞}(2n-1)(2x)^n\]
I know the power series converges for -1/2<x<1/2

Answer 2

The sum function? Sorry, just curious. I know about power series, taylor series and things like that, but guess I haven't quite heard of sum function.

Answer 3

I need to find the sum of the series. Immediately I think I somehow need to use geomatrisk series

Answer 4

Oh, just the sum, gotcha.

Answer 5

:) sorry..

Answer 6

Well, I've stared at this.....and being inexperienced with it, it took too long xD BUT! I have something for ya.

Answer 7

great :)

Answer 8

Alright, so we have to use the interval of convergence to pick out some value. Because what we need to do is use that value to make a partial sum, then turn the partial sum into a series of its own. So within the interval of convergence, I chose x = 1/4. So first I just used n =0, 1, 2, 3, 4 to put the first few terms of our series. Doing this gave me:

\[-1 +2x + 12x ^{2}+40x ^{3}+112x ^{4}.....\]
Plugging in x = (1/4), I come up with:
\[\frac{ 1 }{ 2 }+\frac{ 3 }{ 4 }+\frac{ 5 }{ 8 }+\frac{ 7 }{ 16 }...=\frac{ n+1 }{ 2^{n} }\]
Now I have to somehow make this into a geo-series. One way I can simplify it is split it into 2 fractions and look at it as a limit.
\[\sum_{n=1}^{\infty}\frac{ n }{ 2^{n} }+ \lim_{n \rightarrow \infty }\frac{ 1 }{ 2^{n} }\]
Now the reason I do this is because this shows that as n goes to infinity, the impact of the +1 in the numerator will have no real consequence on the sum. Also notice that the first term when n was 0 was a -1.Essentially this truly cancels out it's significance and I can focus purely on the (n/2^n) series.
Now only using the (n/2^n) series, my sums look like:
\[\frac{ 1 }{ 2 }+\frac{ 2 }{ 4 }+\frac{ 3 }{ 8 }+\frac{ 4 }{ 16 }+\frac{ 5 }{ 32 }...\]Now I wish I had some sort of proof for this, but it was something that I was shown and it seems to provide the right answer for this problem, so yeah. Now the first two terms are basically the same, so I'm going to ignore it, it's kind of an anomaly for this. So what I was shown was to divide the 1st term into the 2nd. If I do that, I have (3/8) / (1/2) = 3/4. So when you do this, the first term, the 1/2, is your a and the result of the division is your r. So then I can use those values to plug them into the formula for the sum of a geo-series:
\[\frac{ \frac{ 1 }{ 2 } }{ 1-\frac{ 3 }{ 4 } } = 2\]
I truly wish I had a better proof for that last part, but I can tell you the limit part is legit and that the sum of n/2^n is truly 2. I just have to go off of something I was vaguely shown and it DOES give the correct answer. Meh, lol.

Answer 9

Great answer. thank you..

Answer 10

Sorry about the wait, lol. I wanted to see if I could manage something out of it and then give an actual explanation that wasn't jumbled thoughts xD

Answer 11

:)