# What is the interval of convergence of sum ( (x-2) ^n ) / ( n^2 + 1 ) ?

Feb 28, 2016

The radius of convergence $R$ of a power series ${\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$ is given by $R = \frac{1}{\lim {\text{sup}}_{n \to \infty} \sqrt[n]{\left\mid {a}_{n} \right\mid}}$.

Here, ${\sum}_{n = 0}^{\infty} \frac{{\left(x - 2\right)}^{n}}{{n}^{2} + 1}$ converges $\forall x \in \left[1 , 3\right] .$

#### Explanation:

First , we have to write ${\sum}_{n = 0}^{\infty} \frac{{\left(x - 2\right)}^{n}}{{n}^{2} + 1}$ in the form of ${\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$. So we will do a change of variable $y = x - 2 \iff x = 2 + y .$

${\sum}_{n = 0}^{\infty} \frac{{\left(x - 2\right)}^{n}}{{n}^{2} + 1} = {\sum}_{n = 0}^{\infty} \frac{{y}^{n}}{{n}^{2} + 1} .$

Therefore, ${a}_{n} = \frac{1}{{n}^{2} + 1} = | {a}_{n} | .$

Now, let's compute $R$ :

$\lim {\text{sup"_(n->oo)root(n)(abs(a_n)) = lim "sup}}_{n \to \infty} \frac{1}{\sqrt[n]{{n}^{2} + 1}} .$

"Yet, " root(n)( n^2 + 1 ) = exp(ln(root(n)( n^2 + 1 ))) = exp(1/nln( n^2 + 1 ))) = e^(ln( n^2 + 1 )/n).

$\text{And " lim_(n->oo)ln( n^2 + 1 )/n = lim_(n->oo) (1/(n^2 + 1) * 2n) / 1 = lim_(n->oo) (2n)/(n^2 + 1) = lim_(n->oo) 2/(2n) = 0 " by Bernouilli l'Hôpital's rule} .$

$\text{Thus, } {\lim}_{n \to \infty} \sqrt[n]{{n}^{2} + 1} = {\lim}_{n \to \infty} {e}^{\ln \frac{{n}^{2} + 1}{n}} = {e}^{0} = 1.$

$\implies R = \lim {\text{sup}}_{n \to \infty} \frac{1}{\sqrt[n]{{n}^{2} + 1}}$

$= {\lim}_{n \to \infty} \frac{1}{\sqrt[n]{{n}^{2} + 1}} = \frac{1}{1} = 1.$

Therefore, ${\sum}_{n = 0}^{\infty} \frac{{y}^{n}}{{n}^{2} + 1}$ converges AA y in ]-1, 1[

$\iff {\sum}_{n = 0}^{\infty} \frac{{\left(x - 2\right)}^{n}}{{n}^{2} + 1}$ converges AA x in ]1, 3[.

Now, we should check the endpoints of the interval.

• If $x = 3$ :

${\sum}_{n = 0}^{\infty} \frac{{1}^{n}}{{n}^{2} + 1} = {\sum}_{n = 0}^{\infty} \frac{1}{{n}^{2} + 1}$ converges because ${n}^{2} + 1 \ge {n}^{2} \forall n > 0 , n \in \mathbb{N} \iff \frac{1}{{n}^{2} + 1} \le \frac{1}{n} ^ 2 \forall n > 0 , n \in \mathbb{N}$ by the comparison test.

• If $x = 1$ :

${\sum}_{n = 0}^{\infty} \frac{{\left(- 1\right)}^{n}}{{n}^{2} + 1}$ converges because it converges absolutely (see the case $x = 3$ above).

$\text{Therefore, } {\sum}_{n = 0}^{\infty} \frac{{\left(x - 2\right)}^{n}}{{n}^{2} + 1}$ converges $\forall x \in \left[1 , 3\right] .$