Show that a convergent sequence is bounded, i.e.: if \(\lim_{ n \to \infty } a_n\) exists, then there is an \(M\) such that \(|a_n| \le M\) for all \(n \ge 1\text{.}\)
Let \((a_n)\) be a sequence. A point \(a\) is an accumulation point of the sequence if for every \(\epsilon > 0\) and every \(N \in \Z_{ >0 }\) there exists some \(n > N\) such that \(|a_n - a| \lt \epsilon\text{.}\) Prove that if a sequence has more than one accumulation point then the sequence diverges.
Prove that, if \(a_n \leq b_n \leq c_n\) for all \(n\) and \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n =
L\text{,}\) then \(\lim_{n \to \infty} b_n = L\text{.}\) This is called the Squeeze Theorem, and is useful in testing a sequence for convergence.
Suppose that the sequence \((c_n)\) converges to zero. Show that \(\sum_{n \ge 0} c_n\) converges if and only if \(\sum_{k \ge 0} (c_{2k}+c_{2k+1})\) converges. Moreover, if the two series converge then they have the same limit.
Suppose \(G\subseteq\C\) and \(f_n : G \to \C\) for \(n \ge 1\text{.}\) Suppose \((a_n)\) is a sequence in \(\R\) with \(\lim_{ n \to \infty } a_n = 0\) and, for each \(n \ge 1\text{,}\)
\begin{equation*}
|f_n(z)| \ \le \ a_n \qquad \text{ for all } z \in G \,\text{.}
\end{equation*}
Show that \((f_n)\) converges uniformly to the zero function in \(G\text{.}\)
Suppose \(G\subseteq\C\text{,}\)\(f_n : G \to \C\) for \(n \ge 1\text{,}\) and \((f_n)\) converges uniformly to the zero function in \(G\text{.}\) Show that, if \((z_n)\) is any sequence in \(G\text{,}\) then
Apply a) to the function sequence given in Example 7.3.2, together with the sequence \((z_n = e^{ - \frac 1 n })\text{,}\) to prove that the convergence given in Example 7.3.2 is not uniform.
Consider \(f_n : [0,\pi] \to \R\) given by \(f_n(x) =
\sin^n(x)\text{,}\) for \(n \ge 1\text{.}\) Prove that \((f_n)\) converges pointwise to \(f: [0,\pi]
\to \R\) given by
\begin{equation*}
f(x) = \begin{cases}1 \amp \text{ if } x = \frac \pi 2 \,
, \\ 0 \amp \text{ if } x \ne \frac \pi 2 \, , \end{cases}
\end{equation*}
yet this convergence is not uniform. (See Figure 7.5.1.)
Treat \(x=0\) as a special case; for \(x>0\) you can use L’Hôpital’s rule (Theorem A.0.11) — but remember that \(n\) is the variable, not \(x\text{.}\)
The product of two power series centered at \(z_0\) is another power series centered at \(z_0\text{.}\) Derive a formula for its coefficients in terms of the coefficients of the original two power series.
Suppose that the sequence \((c_k)\) does not converge to \(0\text{.}\) Show that the radius of convergence of \(\sum_{k\ge0}c_k(z-z_0)^k\) is at most \(1\text{.}\)
Fix \(z \in \C\) and \(r > |z|\text{.}\) Prove that \(\ds \s \left( \frac z w \right)^k\) converges uniformly in the variable\(w\) for \(|w| \ge r\text{.}\)
Prove that, if \(\lim_{ k \to \infty } \left| \frac{ c_{ k+1
} }{ c_{ k } } \right|\) exists then the radius of convergence of \(\s c_k ( z - z_0)^k\) equals
\begin{equation*}
R = \begin{cases}\infty \amp \text{ if } \lim_{ k \to
\infty } \left| \frac{ c_{ k+1 } }{ c_{ k } } \right| = 0 \,
, \\ \lim_{ k \to \infty } \left| \frac{ c_{ k } }{ c_{ k+1
} } \right| \amp \text{ otherwise. } \end{cases}
\end{equation*}
