Prove that the following sum converges and has the given value.
![\[ \sum_{n=1}^{\infty} \frac{2^n + n^2 + n}{2^{n+1} n(n+1)} = 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3611293342d69bc4d25309d5885640ca_l3.png "Rendered by QuickLaTeX.com")
First, we simplify,
![\begin{align*} \sum_{n=1}^{\infty} \frac{2^n + n^2 + n}{2^{n+1} n (n+1)} &= \sum_{n=1}^{\infty} \left( \frac{1}{2n(n+1)} + \frac{1}{2^{n+1}} \right) \\[9pt] &= \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{2^n} \right) \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) + \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{2^n}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5296402f58c205d0d94cb6795448bac0_l3.png "Rendered by QuickLaTeX.com")
Splitting the sum is justified since the two sums both converge since the first one is a telescoping series and the second is a geometric series. Then we have
![\begin{align*} &= \frac{1}{2} (b_1 - L) + \frac{1}{4} \left( \frac{1}{1 - \frac{1}{2}} \right) \\[9pt] &= \frac{1}{2} (1) + \frac{1}{4} \cdot 2 \\[9pt] &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7305df5a48501ad68a7d03f5cfcdf544_l3.png "Rendered by QuickLaTeX.com")