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

\((\frac{5}{2},\frac{7}{2})\)