Just Keep Going

The given sum is :

$S = \displaystyle \sum_{n=0}^{\infty} \dfrac{2n+1}{2^{n}} \\= 2\sum_{n=0}^{\infty} \dfrac{n}{2^{n}} + \sum_{n=0}^{\infty} \dfrac{1}{2^{n}} = 2*S_{1} + S_{2} \quad (say)$

Let's calculate the first sum :

$S_{1} = \sum_{n=0}^{ \infty } x^{n} = \dfrac{1}{1-x} \quad \forall \quad |x| \leq 1$ $\Rightarrow \sum_{n=0}^{ \infty } \left ( \frac{d}{dx} x^{n} \right ) = \frac{d}{dx} \left ( \dfrac{1}{1-x} \right ) \\ \Rightarrow \sum_{n=0}^{ \infty} \left ( n \cdot x^{n} \right ) = \dfrac{x}{ (1-x)^{2} } \\ \Rightarrow \sum_{n=0}^{ \infty} \frac{d}{dx} \left ( n \cdot x^{n} \right ) = \frac{d}{dx} \left ( \dfrac{x}{ (1-x)^{2} } \right )$

Now input $x=\dfrac{1}{2}$ , to get $2*S_{1} = 2*2 = 4$

Now moving on to the second sum :

This is the summation of an infinite GP with common ratio = $\dfrac{1}{2}$

Hence the second sum evaluates to ,

$S_{2} = \dfrac{1}{1-r} = \dfrac{1}{1-\frac{1}{2}} = 2$

So our final answer is 4+2 = 6 .

Q.E.D