Series

Evaluate the series:
$$\mathcal{S}=\sum_{n=1}^{\infty}\frac{2n+1}{2^n}$$

Solution:

 
We are using the formulae:

$ \bullet \displaystyle \;\; \sum_{n=1}^{\infty}x^n =\frac{x}{1-x}, \;\; |x|<1$ which holds since:

$$\begin{aligned}
\sum_{n=0}^{\infty}x^n =\frac{1}{1-x} &\Leftrightarrow 1+ \sum_{n=1}^{\infty}x^n =\frac{1}{1-x}  \\
 &\Leftrightarrow  \sum_{n=1}^{\infty}x^n =\frac{x}{1-x}, \;\; |x|<1\\
\end{aligned}$$

Differentiating the equation above we get that:

$\bullet \displaystyle \;\; \sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{\left ( 1-x \right )^2}, \;\; \left | x \right |<1 $

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Splitting the sum apart (both of the converge the second one as a geometric series and the other one from D'Alembert's test) we have:

$$\sum_{n=1}^{\infty}\frac{2n+1}{2^n}=\sum_{n=1}^{\infty}\left ( \frac{n}{2^{n-1}}+\frac{1}{2^n} \right )=\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}+\require{cancel} \cancelto{1}{\sum_{n=1}^{\infty}\frac{1}{2^n}}$$

Using the second formula we derived from differentiation and plugging in $x=1/2$ we get that first sum equals $4$ hence, the original series is summed up to $5$, that is:

$$\sum_{n=1}^{\infty}\frac{2n+1}{2^n}=5$$

and the exercise comes to an end.