Infi-1 8

\begin{matrix} \sum_{n = 0}^{\infty} x^{n} \to \frac{1}{1 - x} ⟺ | x | < 1 \end{matrix}

Exercise

\begin{matrix} \sum \frac{2^{n} + n^{2} + n}{2^{n + 1} n (n + 1)} \\ a_{n} = \frac{2^{n}}{2^{n + 1} n (n + 1)} + \frac{n^{2} + n}{2^{n + 1} n (n + 1)} \\ a_{n} = \frac{1}{2} \cdot \frac{1}{n (n + 1)} + \frac{1}{2^{n + 1}} \\ \sum \frac{1}{2^{n + 1}} = \sum \frac{1}{4} \cdot {(\frac{1}{2})}^{n - 1} = \sum_{n = 1}^{\infty} \frac{1}{4} \cdot {(\frac{1}{2})}^{n} \to \frac{1}{4} \cdot \frac{1}{1 - \frac{1}{2}} = \frac{1}{2} \\ \frac{1}{n (n + 1)} = \frac{1}{n} - \frac{1}{n + 1} \\ \sum \frac{1}{2} \frac{1}{n (n + 1)} = \frac{1}{2} \sum (\frac{1}{n} - \frac{1}{n + 1}) \to \frac{1}{2} \\ S_{N} = 1 - \frac{1}{N + 1} \to 1 \\ ⟹ \sum \frac{2^{n} + n^{2} + n}{2^{n + 1} n (n + 1)} \to 1 \end{matrix}