If $\sum a_{n} x^{n}$ converges in $( -1,1)$, then $\sum a^{k}_{n} x^{n}$ also converges in $( -1,1)$

Question

I have to prove or disprove the following statement:

If $\sum a_{n} x^{n}$ converges in $( -1,1)$, then $\sum a^{k}_{n} x^{n}$ also converges in $( -1,1)$ , for $k\in \mathbb{N} $.

I actually already proved it. But I'm not 100% sure about my proof. Could someone confirm or find a way to disprove it?

My solution:

$\displaystyle \sum a_{n} x^{n}$ is a power series with center $\displaystyle 0$, it's given that it converges in $\displaystyle ( -1,1)$, from here that the radius is at least $\displaystyle 1$. Let's denote the radius $\displaystyle r$.

By definition $\displaystyle r\ =\ \frac{1}{\lim _{x\rightarrow \infty }\sqrt[n]{|a_{n} |}}$. Since, $\displaystyle 1\leqslant r\leqslant \infty $, it comes that $\displaystyle 0\leqslant \lim _{x\rightarrow \infty }\sqrt[n]{|a_{n} |} \leqslant 1$.

Now let's look at $\displaystyle \sum a^{k}_{n} x^{n}$. Also here we have a power series with center $\displaystyle 0$, in order to prove it converges in $\displaystyle ( -1,1)$ we have to prove the radius is at least $\displaystyle 1$. Let's denote it's radius $\displaystyle r_{k}$. We need to prove that : $\displaystyle 1\leqslant r_{k} \leqslant \infty $, meaning $\displaystyle 1\leqslant \frac{1}{\lim _{x\rightarrow \infty }\sqrt[n]{|a^{k}_{n} |}} \leqslant \infty $, meaning $\displaystyle 0\leqslant \lim _{x\rightarrow \infty }\sqrt[n]{|a^{k}_{n} |} \leqslant 1$.

We know that $\displaystyle 0\leqslant \lim _{x\rightarrow \infty } |a_{n} |^{\frac{1}{n}} \leqslant 1$, it derives that also $\displaystyle 0\leqslant \lim _{x\rightarrow \infty }\left( |a_{n} |^{\frac{1}{n}}\right)^{k} \leqslant 1$, since $\displaystyle k\in \mathbb{N}$.

From here that $\displaystyle 1\leqslant r_{k} \leqslant \infty $, the radius is at least $\displaystyle 1$, and it converges by definition in $\displaystyle ( -1,1)$.

Thanks a lot

Answer 1

My solution:

$\displaystyle \sum a_{n} x^{n}$ is a power series with center $\displaystyle 0$, it's given that it converges in $\displaystyle ( -1,1)$, from here that the radius is at least $\displaystyle 1$. Let's denote the radius $\displaystyle r$.

By definition $\displaystyle r\ =\ \frac{1}{\lim _{x\rightarrow \infty }\sqrt[n]{|a_{n} |}}$. Since, $\displaystyle 1\leqslant r\leqslant \infty $, it comes that $\displaystyle 0\leqslant \lim _{x\rightarrow \infty }\sqrt[n]{|a_{n} |} \leqslant 1$.

Now let's look at $\displaystyle \sum a^{k}_{n} x^{n}$. Also here we have a power series with center $\displaystyle 0$, in order to prove it converges in $\displaystyle ( -1,1)$ we have to prove the radius is at least $\displaystyle 1$. Let's denote it's radius $\displaystyle r_{k}$. We need to prove that : $\displaystyle 1\leqslant r_{k} \leqslant \infty $, meaning \ $\displaystyle 1\leqslant \frac{1}{\lim _{x\rightarrow \infty }\sqrt[n]{|a^{k}_{n} |}} \leqslant \infty $, meaning $\displaystyle 0\leqslant \lim _{x\rightarrow \infty }\sqrt[n]{|a^{k}_{n} |} \leqslant 1$.

We know that $\displaystyle 0\leqslant \lim _{x\rightarrow \infty } |a_{n} |^{\frac{1}{n}} \leqslant 1$, it derives that also $\displaystyle 0\leqslant \lim _{x\rightarrow \infty }\left( |a_{n} |^{\frac{1}{n}}\right)^{k} \leqslant 1$, since $\displaystyle k\in \mathbb{N}$.

From here that $\displaystyle 1\leqslant r_{k} \leqslant \infty $, the radius is at least $\displaystyle 1$, and it converges by definition in $\displaystyle ( -1,1)$.

Answer 2

Let $r$ be the convergence radius of $\sum_n a_nx^n$ and R the convergence raduis of $\sum_na_n^k x^n$, $$\dfrac{1}{R}=\lim_{n\to +\infty}\dfrac{a^k_{n+1}}{a^k_n}=\lim_{n\to +\infty} \dfrac{a_{n+1}}{a_n}=\dfrac{1}{r},$$ so $ r=R$. $\square$

Answer 3

Try ratio test, noting that $|x|<1$:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}^k x^{n+1}}{a_n^k x^n} \right| = |x| \lim_{n\to\infty} \left| \frac{a_{n+1} }{a_n } \right|^k < \lim_{n\to\infty} \left| \frac{a_{n+1} }{a_n } \right|^k \overset{(*)}{<} 1^k=1 $$

Where $(*)$ follows from the fact that the first series is convergent. Since limit of ratio is less than $1$, the series converges