Exercise 6.5.7

Question

Let $\sum a_{n} x^{n}$ be a power series with $a_{n} 
eq 0$, and assume
$$
L=\lim _{n ightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}ight|
$$
exists.
\begin{enumerate}[label=(\alph*)] 
\item Show that if $L 
eq 0$, then the series converges for all $x$ in $(-1 / L, 1 / L)$. (The advice in Exercise 2.7.9 may be helpful.)
\item Show that if $L=0$, then the series converges for all $x \in \mathbf{R}$.
\item Show that (a) and (b) continue to hold if $L$ is replaced by the limit.
$$
L^{\prime}=\lim _{n ightarrow \infty} s_{n} \quad 	ext { where } \quad s_{n}=\sup \left\{\left|\frac{a_{k+1}}{a_{k}}ight|: k \geq night\} .
$$
(General properties of the limit superior are discussed in Exercise 2.4.7.)
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Let $b_n = a_n x^n$. If $|x| < L$, we have $$\lim_{n\to\infty} \left|\frac{b_{n+1}}{b_n}\right| = \lim_{n \to \infty} \left|\frac{a_{n+1}x}{a_n}\right| = \lim_{n \to \infty} |Lx| $$ and thus by the ratio test, if $|Lx| < 1$ then the series $\sum^\infty_{n=1}a_n x^n$ converges. This implies a radius of convergence of $1/L$ if $L \neq 0$. \item By the same logic, if $L = 0$ then $|Lx| < 1$ regardless of $x$ and the series converges $\forall x \in \mathbf{R}$. \item Since $(s_n)$ converges to $L'$, for any $\epsilon > 0$ we have that $\left|a_{k+1} / a_k\right| < M = L' + \epsilon$ once $k > N$ for some $N \in \mathbf{N}$. Therefore by the ratio test and similar logic to above, the radius of convergence is at least $1/M$; since $\epsilon$ is arbitrary, this is effectively a radius of convergence of $1/L$, so (a) and (b) continue to hold. \end{enumerate}

Exercise 6.5.7

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Exercise 6.5.7

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