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 $\sqrt{p}$.

Short 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-by-step solution

Step 1. Given information.

given,

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

Step 2. Consider the power series ∑k=0∞ akxk and ∑k=0∞ akx2k

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 ∑k=0∞ akx2k, 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

localid="1649444846639" $|x{|}^2<\frac{a_k}{a_{k+1}}$

Where, $\sqrt{\frac{a_k}{a_{k+1}}}$is the radius of convergence of the power serieslocalid="1649444777367" $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$

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 $\sqrt{p}$.