Question

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

Answer 1

• The radius of convergence is 2.

,

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

Explanation:

Given the series:

`\sum ^(\infty)_(n=0)(-1)^n(n(x-4)^n)/(2^n)`

We can rewrite it in the form below:

`\sum ^(\infty)_(n=0)(\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)(a_(n+1))/(a_n)=\lim _(n\to\infty)(((-1)^(n+1)(n+1)(x-4)^(n+1))/(2^(n+1)))/(((-1)^nn(x-4)^n)/(2^n))`

First, simplify the fraction:

`\begin{gathered} (((-1)^(n+1)(n+1)(x-4)^(n+1))/(2^(n+1)))/(((-1)^nn(x-4)^n)/(2^n))=((-1)^(n+1)(n+1)(x-4)^(n+1))/(2^(n+1))/((-1)^nn(x-4)^n)/(2^n) \\ =((-1)^(n+1)(n+1)(x-4)^(n+1))/(2^(n+1))*(2^n)/((-1)^nn(x-4)^n) \\ =\frac{(-1)^{}(n+1)(x-4)^{}}{2n} \end{gathered}`

Therefore:

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

In order for the series to converge, we need:

`\begin{gathered} \lim _(n\to\infty)|(a_(n+1))/(a_n)|<1 \\ \implies|(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).