<s>[INST] 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?
 [/INST] 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 $x=1$:
$$\frac{2+1}{(2-1)^2}=3$$
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