Question

# Determine the radius and interval of convergence for each of the following power series. sum_{n=0}^inftyfrac{2^n(x-3)^n}{sqrt{n+3}}

Series
Determine the radius and interval of convergence for each of the following power series.
$$\sum_{n=0}^\infty\frac{2^n(x-3)^n}{\sqrt{n+3}}$$

2021-03-09

Given:
The power series is:
$$\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty\frac{2^n(x-3)^n}{\sqrt{n+3}}$$
We have to determine the radius and interval of convergence for the given power series.
We know that,
When
$$\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|<1$$
Then the power series is converges so we find the interval for x such that the given
power series converges.
We have,
$$a_n=\frac{2^n(x-3)^n}{\sqrt{n+3}}$$ and $$a_{n+1}=\frac{2^{(n+1)}(x-3)^{(n+1)}}{\sqrt{(n+1)+3}}$$
Now,
$$\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|<1$$
$$\lim_{n\to\infty}|\frac{\frac{2^{(n+1)}(x-3)^{(n+1)}}{\sqrt{(n+1)+3}}}{\frac{2^n(x-3)^n}{\sqrt{n+3}}}|<1$$
$$\lim_{n\to\infty}|2(x-3)\frac{\sqrt{n+3}}{\sqrt{n+4}}|<1$$
$$\lim_{n\to\infty}|2(x-3)\frac{\sqrt{1+\frac{3}{n}}}{\sqrt{1+\frac{4}{n}}}|<1$$
$$|2(x-3)|<1$$
$$-1<2(x-3)<1$$
$$-\frac{1}{2}$$
$$-\frac{1}{2}+3$$
$$\frac{5}{2}$$
The interval for the radius for convergence is $$(\frac{5}{2},\frac{7}{2})$$
Now consider,
For $$x=3$$ the radius of convergence is:
$$\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty\frac{2^n(x-3)^n}{\sqrt{n+3}}$$
We get, R is the radius of the convergence which is equal to zero
Hence, the radius of the convergence is zero for $$x=3$$ and the interval for the convergence is
$$(\frac{5}{2},\frac{7}{2})$$