Infinite Geometric Series Issue i have came across a series, i am trying to find its sum knowing the fact that, if it converges and its common ratio ex. r is: -1 < r < 1, then i can use the specified formula $\frac{a}{1-r}$ , which specifically means first term of series over 1 minus common ratio
here is the series 
$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}$
i manipulated it this way to prove its convergence: $\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}(2n-1)\frac{1}{2^n}=\sum_{n=1}^{\infty}(2n-1)\left(\frac{1}{2}\right)^n$
$\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$
using it i get the result 1, which actually should be 3
 A: $$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}\frac{2n}{2^n}-\sum_{n=1}^{\infty}\frac{1}{2^n}=\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-1.$$
We use Maclaurin series for  function $\frac{1}{(1-x)}$
$$\frac{1}{1-x}=1+x+x^2+x^3+\dots = \sum_{n=0}^{\infty} x^{n}$$
Differentiating both sides of  this equation we get that
$$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\dots = \sum_{n=1}^{\infty}n x^{n-1}$$
if $x=\frac 12$ then $\sum_{n=1}^{\infty}n (\frac 12)^{n-1}=\frac{1}{(1-\frac 12)^2}=4$
$$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-1=3$$
A: Since you know it converges, let $S=\sum_{n=1}^\infty\frac{2n-1}{2^n}$. Then we have
$$\begin{align}
S &=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\cdots&\text{so}\\
\frac{1}{2}S&=\phantom{\frac{1}{2}}+\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\cdots&\text{subtracting, we get}\\
S-\frac{1}{2}S&=\frac{1}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\cdots&\text{so}\\
S-\frac{1}{2}S+\frac{1}{2}&=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\cdots&\text{so}\\
 &=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots = 2
\end{align}$$
and thus we have
$$
\frac{1}{2}S+\frac{1}{2}=2
$$
from which we can conclude that $S=3$.
This idiom, suitably manipulating a series as a linear combination leading to known sums, is quite often handy.
A: A helpful trick in such problems is the following: Introduce a variable $x$ to convert your "constant series" into a power series in $x$. This gives access to operations like differentiation or integration which then result in a series you can sum to a simple function you are familiar with. Looking at the given series I'd introduce the as yet unknown function $f$ defined by
$$f'(x)=\sum_{n=1}^\infty {(2n-1)x^{2n-2}\over 2^n}\ .$$
Then, up to an additive constant,
$$f(x)=\sum_{n=1}^\infty{x^{2n-1}\over 2^n}={1\over x}\sum_{n=1}^\infty\left({x^2\over 2}\right)^n={x^2/2\over x(1-x^2/2)}={x\over 2-x^2}\ .$$
It follows that
$$f'(x)={1\over 2-x^2}-{x(-2x)\over(2-x^2)^2}\ ,$$
which implies $f'(1)=3$, and this is your sum.
