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?

Use summation by parts. Given a sequence $f(n)$ define $(\Delta f)(n)=f(n+1)-f(n)$ and note that
$$
\sum_{n=a}^b(\Delta f)(n)=f(b+1)-f(a).
$$
Further given two sequences $f(n)$ and $g(n)$, note that
$$
(\Delta fg)(n)=f(n+1)\Delta g(n)+g(n)\Delta f(n)
$$
whence
$$
\sum_{n=a}^bg(n)\Delta f(n)=[f(b+1)g(b+1)-g(a)f(a)]-\sum_{n=a}^bf(n+1)\Delta g(n)\tag{1}
$$
Let $h(n)=-2/2^n$, and note that $\Delta h(n)=2^{-n}$. So by (1)
$$
\sum_{n=1}^k\frac{2n-1}{2^n}=[(2k+1)(-2^{-k})-1(-1)]+2\sum_{k=1}^n2^{-k}.\tag{2}
$$
Let $k\to \infty$ in (2) and use the formula for a geometric series to deduce that
$$
\sum_{n=1}^\infty\frac{2n-1}{2^n}=1+2=3
$$
