91Ó°ÊÓ

given,

$\underset{k=0}{\overset{\mathrm{∞}}{∑}} \frac{1}{a_k}{x}^k$

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

Therefore, 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 $p$.

Therefore,role="math" localid="1649394184146" $\frac{a_k}{a_k+1}=p$

So, here

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

Since $\frac{a_k}{a_{k+1}}=p$, therefore

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

So according to the ratio test for absolute convergence, the series will converge only when |px|<1

That is,

$\left|x\right|<\frac{1}{p}$

The radius of convergence of the power series $\underset{k=0}{\overset{\mathrm{∞}}{∑}} \frac{1}{a_k}{x}^k$is $\frac{1}{p}$.