Question

Sumfunction of a potensserie

Answer 1

I have this series
\[\sum_{n=0}^{∞}(2n-1)(2x)^n\]
I find the interval of converges with the ratio test to (-1/2,1/2)

Now i need to determine the series sum function at all points in the interval of convergence. can someone help me?

Answer 2

@amistre64 Can you help me again?

Answer 3

split the series into two: \(\sum2n(2x)^n - \sum (2x)^2\).
if you find the formula fr the first one, it's done since the second one is the geometric series.

Answer 4

So if x is between -1/2 and 1/2 the first one will --> 0 when n --> ∞ right?

Answer 5

if \(x\neq 0\) (otherwise the we find the value easily), you can say that \((2x)^n = \frac{(2x)^{n+1}}{2x}\). Use such a trick (here or on the original series) :
\[\sum \frac{\mathrm d g(x)}{\mathrm dx} = \frac{\mathrm d}{\mathrm dx}\sum g(x)\] and you can get through i think

Answer 6

I mean of course if -1/2>x>1/2

Answer 7

on the original series: \[g(x) = (2n-1)(2x)^n = (2n-1) \sqrt{(2x)}^{2n} = (2n-1)\sqrt{2x}^{2n-2}\sqrt{2x}^2\].
and compute \(\frac{\mathrm d}{\mathrm dx} \sqrt{2x}^{2n-1}\) to see what are the missing factors (not depending on \(n\))

Answer 8

"So if x is between -1/2 and 1/2 the first one will --> 0 when n --> ∞ right?"

yes