I am trying to calculate the limit of :
$\lim_{n \to \infty} \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+...+\frac{2n-1}{2^n}$
but I am not sure how to solve it, I thought to calculate of 2S and than subtract S, but it did not worked well. I did noticed that the denominator is a geometric serie,but I dont know how to continue. could you help?
First of all you can easily see that $$\sum_{n\ge 1}\frac{2n-1}{2^n}=\sum_{n\ge 2}\frac{n}{2^{n}}-\sum_{n\ge 1}\frac1{2^n}=4-1=3$$
Than the limit doesn't have sense anymore, because $$\lim_{n \to \infty} \left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+...+\frac{2n-1}{2^n}\right)=\lim_{n \to \infty}\left( \sum_{n\ge 1}\frac{2n-1}{2^n}\right)=\lim_{n \to \infty}3=3$$
If you add $\frac 1{2^i}$ to each term the numerator becomes even. You can then divide out a factor $2$, leaving you with $$\frac 1{2^0}+\frac 2{2^1}+\frac 3{2^2}+\ldots$$. Now you can see this question or this question.
$${ S }_{ n }=\frac { 1 }{ 2 } +\frac { 3 }{ 2^{ 2 } } +\frac { 5 }{ 2^{ 3 } } +...+\frac { 2n-1 }{ 2^{ n } } \\ 2{ S }_{ n }=1+\frac { 3 }{ 2 } +\frac { 5 }{ 2^{ 2 } } +...+\frac { 2n-1 }{ 2^{ n-1 } } \\ 2{ S }_{ n }-{ S }_{ n }=1+\left( \frac { 3 }{ 2 } -\frac { 1 }{ 2 } \right) +\left( \frac { 5 }{ { 2 }^{ 2 } } -\frac { 3 }{ { 2 }^{ 2 } } \right) +...+\left( \frac { 2n-1 }{ { 2 }^{ n-1 } } -\frac { 2n-3 }{ { 2 }^{ n-1 } } \right) -\frac { 2n-1 }{ 2^{ n } } \\ { S }_{ n }=1+1+\frac { 1 }{ 2 } +...+\frac { 1 }{ { 2 }^{ n-2 } } -\frac { 2n-1 }{ { 2 }^{ n } } =1+\frac { 1-\frac { 1 }{ { 2 }^{ n-1 } } }{ 1-\frac { 1 }{ 2 } } -\frac { 2n-1 }{ { 2 }^{ n } } =1+2-\frac { 1 }{ { 2 }^{ n-2 } } -\frac { 2n-1 }{ { 2 }^{ n } } \overset { n\rightarrow \infty }{ \longrightarrow } 3$$
General term of an Arithmetico–geometric sequence is written as, $$a_n = (a + (n-1)d)br^{n-1}$$ and sum up to infinite terms of an AGP is given by
$$S = \dfrac{ab}{1-r} + \dfrac{dbr}{(1-r)^2} \tag 1$$.
First, we can rewrite your sum as
$$S =\lim_{n \to \infty}\sum_{k =1}^n \dfrac{2k-1}{2^k} $$
The general term can be written as $$a_n = (1 + 2(n-1)) \dfrac{1}{2} \times\dfrac{1}{2^{n-1}}$$
You can see that terms in this sum forms an Arithmetico–geometric sequence with $a= 1$, $d = 2$, $b = \dfrac12$ and $r = \dfrac12$.
So the answer is $$S =\dfrac{ab}{1-r} + \dfrac{dbr}{(1-r)^2} = 1+ 2 = 3$$
