Exercise 6.5.5

Question

\begin{enumerate}[label=(\alph*)] 
 \item If $s$ satisfies $0<s<1$, show $n s^{n-1}$ is bounded for all $n \geq 1$.
 \item Given an arbitrary $x \in(-R, R)$, pick $t$ to satisfy $|x|<t<R$. Use this start to construct a proof for Theorem 6.5.6.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Note first that all $ns^{n-1} > 0$, and that for $n + 1 > N > \frac{1}{1-s}$ (with $N \in \mathbf{N}$, we can rearrange for $s$ to have $\frac{n}{n + 1} > s$. This implies that $ns^{n-1} > (n+1) s^n$; thus the sequence in $n$ must be bounded by the maximum of the first $N$ terms. \item Choose $s$ satisfying $|s| = t$ and with $s$ having the same sign as $x$. As a preliminary, note that $\sum^\infty_{n=0}a_n s^{n-1} = (1/s) \sum^\infty_{n=0} a_n s^n$ converges. We have $$\sum^\infty_{n=0} n a_n x^{n-1} = \sum^\infty_{n=0}a_n s^{n-1} n \left(\frac{x}{s}\right)^{n-1} \leq M\sum^\infty_{n=0} a_n s^{n-1}$$ where, denoting $p = x/s$ (with $0 < p < 1$), $M$ is an upper bound for $np^{n-1}$. This completes the proof. \end{enumerate}

Exercise 6.5.5

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Exercise 6.5.5

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