Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$ Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$
I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{2n+1}{2^n} \right)$$ is such that
$$ (\frac{3}{2}-\frac{1}{2})=1, \, (\frac{5}{2^2}-\frac{3}{2^2})=\frac{1}{2},\, \, (\frac{7}{2^3}-\frac{5}{2^3})=\frac{1}{4}\cdots (\frac{2n+1}{2^n}-\frac{2n-1}{2^n})=\frac{1}{2^{n-1}}\text{Which converges to 0 }$$
Too I try use terms of the form $\sum_{n=1}^{\infty}\frac{2n}{2^n}$ and relatione with the orignal sum and  consider the factorization and try sum this kind of terms$$\frac{1}{2}\lim_{n \to \infty }\left(1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}\right)$$.
Update:
I try use partial sum of the form
$$S_1=\frac{1}{2},S_{2}=\frac{5}{2^2},S_{3}=\frac{15}{2^3},S_{4}=\frac{37}{2^4} $$
and try  find $\lim_{n \to \infty }S_{n}$ but I don´t get the term of the numerator.
Unfortunelly I don´t get nice results, I hope someone can give me a idea of how I should start.
 A: Here is a solution with elementary math.
Let's denote
$$S_n = \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$$
You have
$$2S_n = 1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}$$
Then
\begin{align}
S_n = 2S_n - S_n &= 1+\left(\frac{3}{2}-\frac{1}{2} \right)+\ldots+\left(\frac{2n-1}{2^{n-1}} - \frac{2n-3}{2^{n-1}} \right) - \frac{2n-1}{2^{n}}  \\
&= 1+\left(1+\frac{1}{2}+\frac{1}{2^2}\ldots+ \frac{1}{2^{n-2}}  \right) - \frac{2n-1}{2^{n}}  \\
&= 1+2\left(1- \frac{1}{2^{n-1}}  \right) - \frac{2n-1}{2^{n}}  \\
&= 3-\frac{1}{2^{n-2}} - \frac{2n-1}{2^{n}}  \\
\end{align}
Hence,
$$\lim_{n\to\infty} S_n=3$$
A: 1)First you need to proof that this sequence is a absolute convergent sequence.
Hence the convergence properties wont be affected by the finite basic algebraic manipulation ,such as multiplying by a real number ,or exchanging the position of elements of the S(n).


*let S=lim S(n) (n⇒+∞)
1/2 S=1/2 limS(n) (n⇒+∞)

S（n）-1/2S（n）=1/2+1/2+1/2²+...1/2ⁿ⁻¹-（2n+1）/2ⁿ⁺¹
let n⇒+∞ ，THEN $1/2S=1/2+1-0=3/2$
THUS $S=3$
A: 
I hope following answer would help. The given series is an infinite arithmetic-geometric series.
