Q. 68 - 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 $p$, and if m is a positive integer greater than $1$, then the series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{mk}$ has a radius of convergence $\sqrt[m]{p}$.

Step-by-Step Solution

Answer

Ans: It is proved that the radius of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{mk}$ is $\sqrt[m]{p}$

Step 1. Given information.

given,

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

Step 2. For the power series ∑ k = 0 ∞   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 $\left|x\right|<\frac{a_k}{a_{k+1}}$

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

Let us consider the radius of convergence of the power 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 m k &#160; , that is

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

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

Implies that $|x{|}^m<\frac{a_k}{a_{k+1}}$

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

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 $\frac{a_k}{a_{k+1}}=p$, therefore the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{mk}$ is $\sqrt[m]{p}$