91Ó°ÊÓ

given,

$\underset{k=0}{\overset{\mathrm{∞}}{∑}} a_k(x−3{)}^k$

So, the radius of convergence of the series is $2$

Therefore, we try to evaluate the constant term in the power series using the radius of convergence.

Let us consider $b_k=a_k{x}^{k}so{b}_{k+1}=a_{k+1}{x}^{k+1}$

Now apply the ratio test for absolute convergence, that is

$\underset{k→\mathrm{∞}}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k→\mathrm{∞}}{lim} \left|\frac{a_{k+1}}{a_k}x\right|$

So according to the ratio test for absolute convergence, the series will converge only when $\left|\frac{a_{k+1}}{a_k}x\right|<1$

Implies that $\left|x\right|=\frac{a_k}{a_{k+1}}$

where, $\left|x\right|=\frac{a_k}{a_{k+1}}$is the radius of convergence of the power series $\underset{k=0}{\overset{\mathrm{∞}}{∑}} a_k{x}^k$

Since we have already considered the radius of convergence of the power series $\underset{k=0}{\overset{\mathrm{∞}}{∑}} a_k{x}^k$is $2$.

Therefore,

$\frac{a_k}{a_k+1}=2$

Again apply the ratio test. So here,

$\begin{array}{r}\underset{k→\mathrm{∞}}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k→\mathrm{∞}}{lim} \left|\frac{a_{k+1}(x−3{)}^{k+1}}{a_k(x−3{)}^k}\right|\\ =\underset{k→\mathrm{∞}}{lim} \left|\frac{a_{k+1}}{a_k}(x−3)\right|\end{array}$

So according to the ratio test for absolute convergence, the series will converge only when $\left|\frac{a_{k+1}}{a_k}\left(x−x_0\right)\right|<1$

Implies that

$|x−3|<\frac{a_k}{a_{k+1}}$

Plug $\frac{a_k}{a_{k+1}}=2$, from the previous power series

Thus,

$|x-3|<2$

The radius of convergence of the power series $\underset{k=0}{\overset{\mathrm{∞}}{∑}} a_k(x−3{)}^k$is $2$.