To find the radius of convergence of a power series, we can use the ratio test or the root test. For this series, the root test is more straightforward. The radius of convergence \( R \) is given by the formula \( R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}} \), where \( a_n \) is the coefficient of the series. Once we have the radius, we can determine the interval of convergence by checking the endpoints separately.
The given power series is:
\[
\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{n^{2}+1}
\]
This is a power series centered at \(x = 2\).
To find the radius of convergence, we use the Ratio Test. Consider the general term of the series:
\[
a_n = \frac{(x-2)^{n}}{n^{2}+1}
\]
The Ratio Test involves computing the limit:
\[
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]
Substitute \(a_n\) and \(a_{n+1}\):
\[
a_{n+1} = \frac{(x-2)^{n+1}}{(n+1)^{2}+1}
\]
\[
\frac{a_{n+1}}{a_n} = \frac{(x-2)^{n+1}}{(n+1)^{2}+1} \cdot \frac{n^{2}+1}{(x-2)^{n}}
\]
Simplify the expression:
\[
= \frac{(x-2) \cdot (n^{2}+1)}{(n+1)^{2}+1}
\]
Take the absolute value and the limit as \(n \to \infty\):
\[
L = \lim_{n \to \infty} \left| (x-2) \cdot \frac{n^{2}+1}{(n+1)^{2}+1} \right|
\]
Simplify the limit:
\[
L = \left| x-2 \right| \cdot \lim_{n \to \infty} \frac{n^{2}+1}{(n+1)^{2}+1}
\]
The dominant terms in the numerator and denominator are \(n^2\) and \((n+1)^2\), respectively:
\[
= \left| x-2 \right| \cdot \lim_{n \to \infty} \frac{n^{2}}{n^{2} + 2n + 2}
\]
\[
= \left| x-2 \right| \cdot \lim_{n \to \infty} \frac{1}{1 + \frac{2}{n} + \frac{2}{n^2}}
\]
As \(n \to \infty\), the terms \(\frac{2}{n}\) and \(\frac{2}{n^2}\) approach zero:
\[
= \left| x-2 \right| \cdot 1 = \left| x-2 \right|
\]
For the series to converge, the limit \(L\) must be less than 1:
\[
\left| x-2 \right| < 1
\]
Thus, the radius of convergence \(R\) is 1.
The interval of convergence is determined by solving:
\[
1 < x-2 < 1
\]
This simplifies to:
\[
1 < x < 3
\]
We need to check the endpoints \(x = 1\) and \(x = 3\) separately.
Substitute \(x = 1\) into the series:
\[
\sum_{n=1}^{\infty} \frac{(1-2)^{n}}{n^{2}+1} = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}+1}
\]
This is an alternating series. By the Alternating Series Test, it converges because \(\frac{1}{n^{2}+1}\) is decreasing and approaches zero.
Substitute \(x = 3\) into the series:
\[
\sum_{n=1}^{\infty} \frac{(3-2)^{n}}{n^{2}+1} = \sum_{n=1}^{\infty} \frac{1}{n^{2}+1}
\]
This series converges by the p-series test with \(p = 2 > 1\).
The radius of convergence is \(R = 1\), and the interval of convergence is:
\[
\boxed{[1, 3]}
\]
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