Interval of Convergence #001

First, perform the ratio test on the function:

$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$ $L = \lim_{n \to \infty} \left| \frac{(-1)^{n+1}(n+1)(x+3)^{n+1}}{4^{n+1}} \cdot \frac{4^n}{(-1)^nn(x+3)^{n}} \right|$

$\therefore L = \lim_{n \to \infty} \left| \frac{(-1)(n+1)(x+3)}{4n} \right|$

$L = \lim_{n \to \infty} \left| \frac{(n+1)(x+3)}{4n} \right|$ $L = \left| \frac{x+3}{4} \right| \lim_{n \to \infty} \left| \frac{(n+1)}{n} \right|$ $\therefore L = \frac{\left|x+3\right|}{4}$

In order for the series to converge, L must be less than 1, so:

$\frac{\left|x+3\right|}{4} < 1$ $\left|x+3\right| < 4$

The Radius of convergence for the series is, therefore, equal to 4. To find the Interval of convergence, solve the inequality above:

$-4 < x+3 < 4$ $-7 < x < 1$

However, this is not necessarily the answer. We need to check whether the boundary values converge or diverge. First, substituting the value $x=-7$ in to the series:

$\sum_{n=0}^{\infty} \frac{(-1)^nn(-7+3)^n}{4^n}$ $=\sum_{n=0}^{\infty} \frac{(-1)^nn(-4)^n}{4^n}$ $=\sum_{n=0}^{\infty} (-1)^n(-1)^nn$ $=\sum_{n=0}^{\infty} n$

which does not converge. Substituting in for $x=1$ :

$\sum_{n=0}^{\infty} \frac{(-1)^nn(1+3)^n}{4^n}$ $=\sum_{n=0}^{\infty} \frac{(-1)^nn(4)^n}{4^n}$ $=\sum_{n=0}^{\infty} (-1)^nn$

which, by the Divergence Test, also does not converge. So neither of the boundary values are included in the interval.

Thus the answer is $\boxed{-7 < x < 1}$ .

Consider the following sum:

$\begin{aligned} \sum_{n=0}^\infty (-1)^n u^n & = \frac 1{1+u} & \small \color{#D61F06} \text{for }-1< u < 1 \\ \sum_{n=0}^\infty (-1)^n nu^{n-1} & = - \frac 1{(1+u)^2} & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }u. \\ \sum_{n=0}^\infty (-1)^n nu^n & = - \frac u{(1+u)^2} & \small \color{#3D99F6} \text{Multiply both sides by }u. \\ \sum_{n=0}^\infty \frac {(-1)^n n(x+3)^n}{4^n} & = - \frac {4(x+3)}{(x+7)^2} & \small \color{#3D99F6} \text{Put }u = \frac {x+3}4 \end{aligned}$

We note that the sum converges only when

$\begin{array} {lc} &\color{#D61F06} - 1 < u < 1 \\ & - 1 < \dfrac {x+3}4 < 1 \\ \implies & \boxed{- 7 < x < 1} \end{array}$