Find the limit of $\lim_{n\rightarrow\infty}(\frac{1}{2}+\frac{3}{2^2}+...+\frac{2n-1}{2^n})$ Can anybody help me with this limit? 
$$\lim_{n\rightarrow\infty}\left(\frac{1}{2}+\frac{3}{2^2}+\cdots+\frac{2n-1}{2^n}\right)$$
Abel's transform wouldn't work on this one I guess, because another serie diverges
As this contains a geometric series with ratio $1/2$, I had multiplied whole series by $1/2$ and then subtracted from original, but got original series back. Any ideas how can the limit be calculated?
 A: $S \ \ = \ \ \ \ \ \ \ \dfrac{1}{2}+\dfrac{3}{2^2}+\dfrac{5}{2^3}+\dfrac{7}{2^4}+\cdots$. 
$2S = 1+\dfrac{3}{2}+\dfrac{5}{2^2}+\dfrac{7}{2^3} + \dfrac{9}{2^4} + \cdots$
Now, subtract the first equation from the second, but match up terms with the same denominator: 
$S = 1+ \dfrac{2}{2} + \dfrac{2}{2^2} + \dfrac{2}{2^3} + \cdots$
This becomes $1$ plus a geometric series, which is easy to evaluate. 
A: We have:
$$\lim_{n\to\infty}\left(\frac{1}{2}+\frac{3}{2^2}+\cdots+\frac{2n-1}{2^n}\right) = \sum_{n=1}^{\infty}\frac{2n-1}{2^n} = \sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-\sum_{n=1}^{\infty}\frac{1}{2^n}.$$
The sum on the right is a standard geometric series. The sum on the left can be determined by taking the usual geometric series:
$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n,\quad |x|<1$$
differentiating both sides and multiplying by $x$ after. I can expand further if you get stuck.
A: prove with induction that $$\sum_{i=1}^{n}\frac{2i-1}{2^i}=-2\, \left( 1/2 \right) ^{n+1}-4\, \left( 1/2 \right) ^{n+1} \left( n+
1 \right) +3
$$
A: You can generalize it to:
\begin{align*}\sum_{n=0}^\infty\frac {2n+1}{2^{n+1}}x^{2n}&=\left(\sum_{n=0}^\infty\frac 1{2^{n+1}}x^{2n+1}\right)'=\left(\frac x2\sum_{n=0}^\infty\frac{x^{2n}}{2^n}\right)'=\left(\frac x2\cdot\frac 1{1-x^2/2}\right)'=\left(\frac x{2-x^2}\right)'\\&=\frac{2-x^2+2x^2}{(2-x^2)^2}=\frac{2+x^2}{(2-x^2)^2}\end{align*}
This holds for $|x^2/2|<1\iff |x|<\sqrt2$, so you can get the answer by setting $x=1$:
$$\frac{2+1}{(2-1)^2}=3$$
A: In the same spirit as user2345215's answer, let us consider $$S=\sum_{i=1}^{\infty}(2i-1)x^i=2x\sum_{i=1}^{\infty}ix^{i-1}-\sum_{i=1}^{\infty}x^{i}=2x\frac{d}{dx}\Big(\sum_{i=1}^{\infty}x^{i}\Big)-\sum_{i=1}^{\infty}x^{i}$$ and $$\sum_{i=1}^{\infty}x^{i}=-1+\sum_{i=0}^{\infty}x^{i}=-1+\frac{1}{1-x}=\frac{x}{1-x}$$ So, compute the derivative, get the expression and replace $x$ by $\frac 12$.
