How can I evaluate and prove $$\sum_{n=1}^\infty \frac{2n-1}{2^n}= 3$$
and how to evaluate and prove $$\sum_{n=1}^\infty \frac{n}{2^n}= 2$$
? I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.
In general, how can I evaluate,please give me the answer step by step with explain please and thanks for every thing
Here's a clever little trick $$S=\sum_{n=1}^\infty\frac{n}{2^n}=\sum_{n=0}^\infty\frac{n+1}{2^{n+1}}=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^\infty\frac{n+1}{2^n}=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^\infty\frac{n}{2^n}+\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2^n}=1+\frac{S}{2}$$ $$S=1+\frac{S}{2}$$ $$S=2$$
Let $$S=\sum_{n=1}^\infty \frac{n}{2^n}$$ Consider the geometric series: $$\sum_{n=1}^\infty x^n =\frac{x}{1-x}\quad\text{if}\quad |x|<1$$ Now take the derivative: $$\frac{d}{dx}\left(\sum_{n=1}^\infty x^n \right)=\frac{d}{dx}\left(\frac{x}{1-x}\right)=\frac{1}{(1-x)^2} $$ Since you are dealing with a power series, you can swap $\Sigma$ and $\frac{d}{dx}$: $$\frac{d}{dx}\left(\sum_{n=1}^\infty x^n \right)=\sum_{n=1}^\infty \frac{d}{dx} (x^n)=\sum_{n=1}^\infty nx^{n-1}$$ From this chain of equality you obtain: $$\frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1} $$ In this particular case, $x=1/2\neq0$, so you can multiply both sides by $x$: $$\frac{x}{(1-x)^2}=\sum_{n=1}^\infty nx^n $$ Plugging $x=1/2$ in the final formula, you obtain $S=2$.
Hint:
Let $\dfrac{2n-1}{2^n}=f(n)-f(n-1)$ where $f(m)=\dfrac{am+b}{2^m}$
$$\implies\dfrac{2n-1}{2^n}=\dfrac{an+b-2(a(n-1)+b)}{2^n}\implies2n-1=-2an+2a-b$$
Comparing the coefficients of $n$ and the constants $$-2a=2\iff a=?, 2a-b=-1\iff b=2a+1=?$$
Now, $$\sum_{n=1}^r\dfrac{2r-1}{2^r}=\sum_{n=1}^r(f(n)-f(n-1))=f(r)-f(0)$$
