Prove that if the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$and $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$have the same radius of convergence $p$, then $p$is $0,1,$ or infinite.

given,

$\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$and$\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$

Also, 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 localid="1649479430552" role="math" $\left|\frac{a_{k+1}}{a_k}x\right|<1$

Implies thatlocalid="1649479926176" role="math" $\left|x\right|<\frac{a_k}{a_{k+1}}$

Where,$\frac{a_k}{a_{k+1}}$is the radius of convergence of the series localid="1649445916913" $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$

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

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

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, role="math" localid="1649446426171" $\sqrt{\frac{a_k}{a_{k+1}}}$is the radius of convergence of the power series role="math" localid="1649446374085" $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$

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

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

Hence, the only solution to the equations$p=\sqrt{p}$are$0,1and\infty$