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.

Using the identity $$ \frac{2k+1}{2^{k-1}}-\frac{2k+3}{2^k}=\frac{2k-1}{2^k}\tag1 $$ the sum can be written as a telescoping series $$ \begin{align} \sum_{k=1}^\infty\frac{2k-1}{2^k} &=\sum_{k=1}^\infty\left(\frac{2k+1}{2^{k-1}}-\frac{2k+3}{2^k}\right)\\ &=3\tag2 \end{align} $$

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.

$$\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}{2^{n}}\right)=\sum _1^{\infty} \frac {2n-1}{2^n}=2\sum _1^{\infty} \frac {n}{2^n} -\sum _1^{\infty} \frac {1}{2^n}=2\sum _1^{\infty} \frac {n}{2^n} -1$$

Note that

\begin{align} \sum _1^{\infty} \frac {n}{2^n}&=\left(\frac12+\frac14+\frac14+\frac18+\frac18+\frac18+\cdots\right)\\&=\left(\frac12+\frac14+\frac18 +\cdots\right) +\left(\frac14+\frac18+\frac1{16}+\cdots\right)+\left(\frac18+\frac1{16}+\frac1{32}+\cdots\right)+\cdots\\&=1+\frac12+\frac14+\frac18+\cdots=2 \end{align}

Thus we have $$\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)= 3$$