Question

I need help with the second problem would you please help understand it??

Answer 1

Answer:

• The radius of convergence is 2.

,

• The interval of convergence is (2,6).

Explanation:

Given the series:

$\sum ^{\infty}_{n=0}(-1)^n\frac{n(x-4)^n}{2^n}$

We can rewrite it in the form below:

$\sum ^{\infty}_{n=0}\frac{\mleft(-1\mright)^nn(x-4)^n}{2^n}$

We apply the ratio's test to find the radius of convergence:

$\lim _{n\to\infty}\frac{a_{n+1}}{a_n}=\lim _{n\to\infty}\frac{\frac{(-1)^{n+1}(n+1)(x-4)^{n+1}}{2^{n+1}}}{\frac{(-1)^nn(x-4)^n}{2^n}}$

First, simplify the fraction:

$\begin{gathered} \frac{\frac{(-1)^{n+1}(n+1)(x-4)^{n+1}}{2^{n+1}}}{\frac{(-1)^nn(x-4)^n}{2^n}}=\frac{(-1)^{n+1}(n+1)(x-4)^{n+1}}{2^{n+1}}\div\frac{(-1)^nn(x-4)^n}{2^n} \\ =\frac{(-1)^{n+1}(n+1)(x-4)^{n+1}}{2^{n+1}}\times\frac{2^n}{(-1)^nn(x-4)^n} \\ =\frac{(-1)^{}(n+1)(x-4)^{}}{2n} \end{gathered}$

Therefore:

$\begin{gathered} \implies\lim _{n\to\infty}\frac{\frac{(-1)^{n+1}(n+1)(x-4)^{n+1}}{2^{n+1}}}{\frac{(-1)^nn(x-4)^n}{2^n}}=\frac{-(x-4)}{2}\lim _{n\to\infty}\frac{(n+1)^{}}{n} \\ \text{Divide all though by n} \\ =\frac{-(x-4)}{2}\lim _{n\to\infty}\frac{(\frac{n}{n}+\frac{1}{n})^{}}{\frac{n}{n}} \\ =\frac{-(x-4)}{2}\lim _{n\to\infty}(1+\frac{1}{n})where\begin{cases}\lim _{n\to\infty}(1)=1 \\ \lim _{n\to\infty}(\frac{1}{n})=0\end{cases} \\ \text{Therefore, the limit is:} \\ \lim _{n\to\infty}\frac{\frac{(-1)^{n+1}(n+1)(x-4)^{n+1}}{2^{n+1}}}{\frac{(-1)^nn(x-4)^n}{2^n}}=\frac{-(x-4)}{2} \end{gathered}$

In order for the series to converge, we need:

$\begin{gathered} \lim _{n\to\infty}|\frac{a_{n+1}}{a_n}|<1 \\ \implies|\frac{x-4}{2}|<1 \\ \implies|x-4|<2 \end{gathered}$

The radius of convergence is 2.

[tex]\begin{gathered} |x-4|<2 \\ -2The interval of convergence is (2,6).