Q. 66 - Calculus

Question

Prove that if the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ has a positive and finite radius of convergence $p$, then the series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$ has a radius of convergence $\sqrt{p}$.

Step-by-Step Solution

Answer

Ans:  Since we have already considered the radius of convergence of the series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ is $p$, that is $\frac{a_k}{a_{k+1}}=p$, the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$ is $\sqrt{p}$

Step 1. Given information.

given,

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

Step 2. Consider the power series ∑ k = 0 ∞   a k x k and ∑ k = 0 ∞   a k x 2 k

 For the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$, let us consider $b_k=a_k{x}^k,\text{ so }{b}_{k+1}=a_{k+1}{x}^{k+1}$

Apply the ratio test for absolute convergence in the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$, 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, $\frac{a_k}{a_{k+1}}$ is the radius of convergence of the series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$.

Let us now consider the radius of convergence of the series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$  is $p$, therefore $\frac{a_k}{a_{k+1}}=p$

Step 3. Again, we apply the ratio test for absolute convergence in the power series &#8721; k = 0 &#8734; &#8202; a k x 2 k , that is

 $\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}}{a_k}{x}^2\right|$

So according to the ratio test for absolute convergence, the series will converge only 

    $\left|\frac{a_{k+1}}{a_k}{x}^2\right|<1$

Implies that

    $|x{|}^2<\frac{a_k}{a_{k+1}}$

Where, $\sqrt{\frac{a_k}{a_{k+1}}}$ is the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$

Step 4. Thus,

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

is $p$, that is $\frac{a_k}{a_{k+1}}=p$, the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$ is $\sqrt{p}$.