Compute the sum of the series ∑ (-2)n ((n+2) / (n+1)) xn

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} (-2)^n \frac{n+2}{n+1} x^n$

converges and compute the sum.

First, we apply the ratio test,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(-2)^{n+1}(n+3)x^{n+1}}{n+2} \right) \left( \frac{n+1}{(-2)^n (n+2) x^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{-2x (n+1)(n+3)}{(n+2)^2} \\[9pt] &= -2x. \end{align*}$

Thus, the series converges if $|-2x| < 1$ which implies $|x| < \frac{1}{2}$ . For the boundary, $|x| = \frac{1}{2}$ we have in the case $x = \frac{1}{2}$ ,

$\sum_{n=0}^{\infty} (-2)^n \frac{n+2}{n+1} x^n = \sum_{n=0}^{\infty} (-1)^n \frac{n+2}{n+1}$

which diverges since the terms do not go to 0. Similarly, if $x = -\frac{1}{2}$ we get $\sum \frac{n+2}{n+1}$ which also diverges since the terms do not go to 0. Therefore, the given series converges absolutely on the interval $|x| < \frac{1}{2}$ . Then, to compute the sum for $|x| < \frac{1}{2}$ we have

$\begin{align*} \sum_{n=0}^{\infty} \frac{(-2x)^n (n+2)}{n+1} &= \sum_{n=0}^{\infty} \frac{(-2x)^n (n+1+1)}{n+1} \\[9pt] &= \sum_{n=0}^{\infty} \frac{(-2x)^n (n+1)}{n+1} + \sum_{n=0}^{\infty} \frac{(-2x)^n}{n+1} &(\text{Abs. Conv. Series}) \\[9pt] &= \sum_{n=0}^{\infty} (-2x)^n + \frac{1}{2x} \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{n+1} \\[9pt] &= \frac{1}{1+2x} + \frac{\log (1+2x)}{2x}. \end{align*}$

Stumbling Robot

Compute the sum of the series ∑ (-2)ⁿ ((n+2) / (n+1)) xⁿ

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