Question

Please compute the series
\[\sum\limits_{n = 1}^\infty {\frac{{2n - 1}}
{{{2^n}}}} \]

Answer 1

I have the answer from the text book but I don't know why...The answer is 3.

Answer 2

I'm trying to remember how to do these but I keep hitting walls. The farthest I'm getting is \[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{2i-1}{2^{i}} = \lim_{n \rightarrow \infty} \frac{n^{2} + n-1}{2^{\frac{n^{2}+n}{2}}}\]

Answer 3

that's with n=0, but the question starts with n=1

Answer 4

lols im sorry :S

Answer 5

The series can be written as:
\[2\sum_{n=1}^{\infty}{n \over 2^n}-\sum_{n=1}^{\infty}{1 \over 2^n}\]
The second term \(-\sum_{n=1}^{\infty}{1 \over 2^n}\) is nothing but a geometric series and can be written as \[-{1 \over 2}\sum_{n=1}^{\infty}({1 \over 2})^{n-1}\]. It has a sum of \(-\frac{1}{2}\times \frac{1}{1-\frac{1}{2}}=-1\).
I am a bit not so sure about how to do the first term, but I'll look into it.

Answer 6

it is very easy :
\[\sum_{n=1}^{\infty}(2n-1)\div 2^n =\sum_{n=1}^{\infty} (2n-1)*(.5)^n\]
which is the Arithmetic-geometric series which has the form
\[\sum_{n=1}^{\infty}(a+nd)*r^n\]
with a=-1 , d=2 ,r=.5
|r|<1 then the sum equals
\[Sum=a/(1-r)+rd/(1-r)^2\]
Sum=-1/.5+1/.5^2
= -2+4
= 2

Answer 7

\[\sum\limits_{n = 1}^\infty {nd*{r^n} = \frac{{rd}}
{{(1 - r)}}} \]

I haven't learn it yet. Let me check..

Answer 8

Could you give me a link ?

Answer 9

this image is from "Mathematical methods for physicists and scientists" Book