Question

Let $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$be a power series in x with an interval of convergence$[-2,2)$. What is the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k(x-3{)}^k$? Justify your answer.

Short answer

Ans: The radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k(x-3{)}^k$is$2$.

Step-by-step solution

Step 1. Given information. 

given,

$\sum _{k=0}^{\mathrm{\infty }} a_k(x-3{)}^k$

Step 2. Solution:

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

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

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 $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$

Since we have already considered the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$is $2$.

Therefore,

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

Step 3. Now, to find the radius of convergence of the power series ∑k=0∞ ak(x−3)k

Again apply the ratio test. So here,

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

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,

$|x-3|<2$

Step 4. Therefore,

The radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k(x-3{)}^k$is $2$.