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}}

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asked 2021-03-08
Determine the radius and interval of convergence for each of the following power series.
\(\sum_{n=0}^\infty\frac{2^n(x-3)^n}{\sqrt{n+3}}\)

Expert Answers (1)

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

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