# Sequences and Series

## Sequences of functions

For $$f:A\to {\mathbf{R}}$$, \begin{align*} \limsup_{x\to y} f(x) \coloneqq\lim_{{\varepsilon}\to 0} \sup f\qty{A \cap B_{\varepsilon}(y) \setminus\left\{{ y }\right\}} .\end{align*}

A series of continuous functions that does not converge uniformly but is still continuous: \begin{align*} g(x) \coloneqq\sum {1 \over 1 + n^2 x} .\end{align*}

Take $$x = 1/n^2$$.

## Sequences of number

$$\limsup$$ is largest limit of a convergent subsequence, $$\liminf$$ is the smallest.

For $$\left\{{a_k}\right\}$$ is a non-increasing sequence in $${\mathbf{R}}$$ then \begin{align*} \sum_{k\geq 1} a_k < \infty \iff \sum_{k\geq 1} 2^k a_{2^k}<\infty .\end{align*}

Show that \begin{align*} \sum a_k \leq \sum 2^k a_{2^k} \leq 2 \sum a_k \end{align*} using \begin{align*} \sum a_k = a_0 + a_1 + a_2 + a_3 + \cdots \leq \qty{a_1} + \qty{a_2 + a_2} + \qty {a_3 + a_3 + a_3 + a_3} + \cdots \\ \end{align*} where each group with $$a_k$$ has $$2^k$$ terms.

## Series

If $$0\leq a_n \leq b_n$$, then

• $$\sum b_n < \infty \implies \sum a_n < \infty$$, and
• $$\sum a_n = \infty \implies \sum b_n = \infty$$.

Given a point $$c$$ and some $$\varepsilon>0$$, if $$f \in C^\infty(I)$$ and there exists an $$M$$ such that \begin{align*} x \in N_\varepsilon(c) \implies {\left\lvert {f^{(n)}(x)} \right\rvert} \leq M^n \end{align*} then the Taylor expansion about $$c$$ converges on $$N_\varepsilon(c)$$.

Let $$n$$ be a fixed dimension and set $$B = \left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}{\left\lVert {x} \right\rVert} \leq 1}\right\}$$. \begin{align*} \sum \frac 1 {n^p} < \infty &\iff p>1 \\ \int_\varepsilon^\infty \frac 1 {x^p} < \infty &\iff p>1 \\ \int_0^1 \frac 1 {x^p} < \infty &\iff p<1 \\ \int_B \frac{1}{{\left\lvert {x} \right\rvert}^p} < \infty &\iff p < n \\ \int_{B^c} \frac{1}{{\left\lvert {x} \right\rvert}^p} < \infty &\iff p > n \\ .\end{align*}

\begin{align*} \sum f_n < \infty \implies {\left\lVert {f_n} \right\rVert}_\infty \overset{n\to\infty}\longrightarrow 0 .\end{align*}

\begin{align*} f_n \text{ cts},\, {\left\lVert {\sum_{k\leq N} f_n \to F} \right\rVert}_\infty \overset{N\to\infty}\longrightarrow 0 \implies F \text{ cts} .\end{align*}

Uniformly Cauchy iff uniformly convergent, i.e. \begin{align*} {\left\lVert {f_n - f_m} \right\rVert} \overset{m, n\to \infty}\longrightarrow 0 \iff \exists f,\, {\left\lVert {f_n - f} \right\rVert} \overset{n\to\infty}\longrightarrow 0 .\end{align*}

If $$f$$ is $$n$$ times differentiable on a neighborhood of a point $$p$$, say $$N_\delta(p)$$, then for all points $$x$$ in the deleted neighborhood $$N_\delta(p) - \left\{{p}\right\}$$ , there exists a point $$\xi$$ strictly between $$x$$ and $$p$$ such that \begin{align*} x \in N_\delta(p)-\left\{{p}\right\} \implies f(x) &= \sum_{k=0}^{n-1} \frac{f^{(k)}(p)}{k!}(x-p)^k + \frac{f^{(n)}(\xi)}{n!}(x-p)^n \\ \\ &= \sum_{k=0}^{n-1} \frac{f^{(k)}(p)}{k!}(x-p)^k + \int_c^x \frac{1}{n!} {\frac{\partial ^n f}{\partial x^n}\,}(t) (x-t)^n ~dt \end{align*}

## Uniform Convergence

$$f_n \to f$$ uniformly iff there exists an $$M_n$$ such that $${\left\lVert {f_n - f} \right\rVert}_\infty \leq M_n \to 0$$.

Negating: find an $$x$$ which depends on $$n$$ for which $${\left\lVert {f_n} \right\rVert}_\infty > {\varepsilon}$$ (negating small tails) or $${\left\lVert {f_n - f_m} \right\rVert} > {\varepsilon}$$ (negating the Cauchy criterion).

The space $$X = C([0, 1])$$, continuous functions $$f: [0, 1] \to {\mathbf{R}}$$, equipped with the norm \begin{align*} {\left\lVert {f} \right\rVert}_\infty \coloneqq\sup_{x\in [0, 1]} {\left\lvert {f(x)} \right\rvert} \end{align*} is a complete metric space.

• Let $$\left\{{f_k}\right\}$$ be Cauchy in $$X$$.

• Define a candidate limit using pointwise convergence:

Fix an $$x$$; since \begin{align*} {\left\lvert {f_k(x) - f_j(x)} \right\rvert} \leq {\left\lVert {f_k - f_k} \right\rVert} \to 0 \end{align*} the sequence $$\left\{{f_k(x)}\right\}$$ is Cauchy in $${\mathbf{R}}$$. So define $$f(x) \coloneqq\lim_k f_k(x)$$.

• Show that $${\left\lVert {f_k - f} \right\rVert} \to 0$$: \begin{align*} {\left\lvert {f_k(x) - f_j(x)} \right\rvert} < \varepsilon ~\forall x \implies \lim_{j} {\left\lvert {f_k(x) - f_j(x)} \right\rvert} <\varepsilon ~\forall x \end{align*} Alternatively, $${\left\lVert {f_k-f} \right\rVert} \leq {\left\lVert {f_k - f_N} \right\rVert} + {\left\lVert {f_N - f_j} \right\rVert}$$, where $$N, j$$ can be chosen large enough to bound each term by $$\varepsilon/2$$.

• Show that $$f\in X$$:

The uniform limit of continuous functions is continuous.

In other cases, you may need to show the limit is bounded, or has bounded derivative, or whatever other conditions define $$X$$.

If $$[a, b] \subset {\mathbf{R}}$$ is a closed interval and $$f$$ is continuous, then for every $${\varepsilon}> 0$$ there exists a polynomial $$p_{\varepsilon}$$ such that $${\left\lVert {f- p_{\varepsilon}} \right\rVert}_{L^\infty([a, b])} \overset{{\varepsilon}\to 0}\to 0$$.

Equivalently, polynomials are dense in the Banach space $$C([0, 1], {\left\lVert {{-}} \right\rVert}_\infty)$$.