<s>[INST] Find the limit of sequence $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}$ without using of derivatives and etc. I need to find limit of sequence
$$
\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)
$$
I tried to solve it and stopped here
$$
f(n+1) = \frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}+\frac{2n+1}{2^{n+1}}
$$
$$
2f(n+1) = 1+\frac{3}{2}+\frac{5}{2^{2}}+\cdots+\frac{2n-1}{2^{n-1}}+\frac{2n+1}{2^{n}}
$$
$$
2f(n+1) -f(n) = 1+ \left(1 + \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\right) = 1 + g(n)
$$
I can find the limit of $g$, but what to do with the other parts?
 [/INST] Let's assume that the limit exists. If we call the limit $s$ then
$s = \sum_1^{\infty}\frac{2n-1}{2^n}$
$\Rightarrow 2s = 1 + \sum_1^{\infty}\frac{2n+1}{2^n} = 1 + \sum_1^{\infty}\frac{2n-1}{2^n} + \sum_1^{\infty}\frac{2}{2^n}$
$\Rightarrow 2s = 1 + s + \sum_0^{\infty}\frac{1}{2^n}$
$\Rightarrow 2s = 1 + s + 2$
$\Rightarrow s=3$
So if the limit exists then it must be 3.
Now you just have to prove that the limit exists i.e. the series converges.
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