2. Sequences

Convergence of sequences via ε-N arguments: monotone and Cauchy sequences, limsup and liminf.

• The ε–N Definition — Watch the moment the whole tail of a sequence fits inside the $\varepsilon$-band. That $N$ is exactly the $N$ in $\forall\varepsilon>0\ \exists N$.
• Sequence and Function Spaces — $\ell^1, \ell^2, \ell^\infty$ and $C[0,1]$ — vector spaces under pointwise operations, metric spaces via the norm $\|x - y\|$. A normed space bridges both worlds. Watch the inclusion chain $\ell^1 \subset \ell^2 \subset \ell^\infty$ through $x_n = 1/n^\alpha$. Completeness of $\ell^p$ and $C[0,1]$ (Banach spaces) is stated as a known theorem only.
• Cauchy–Schwarz Inequality — $|\langle x,y\rangle| \le \|x\|\,\|y\|$ — the inner product never exceeds the product of the norms. Proved from the discriminant of $0 \le \|x-ty\|^2$; equality holds exactly for parallel vectors.
• Why ℝ Has No Holes — Every term of $x_{n+1}=\frac{x_n+2/x_n}{2}$ is rational, yet the limit $\sqrt{2}$ is not. The rationals $\mathbb{Q}$ have a hole — shown at the idea level, without a full irrationality proof.
• Construction of the Real Numbers — A real number is an equivalence class of Cauchy sequences of rationals. Lengthening the decimal expansion of $\sqrt{2}$ marches rational terms into the hole — the march itself is the real number $\sqrt{2}$. Completeness is what this construction buys.
• A Convergent Subsequence Always Hides Inside — Every bounded sequence hides a convergent subsequence. Halve the interval and follow the half holding infinitely many terms.
• Monotone and Cauchy Sequences — $a_n = 1 - 2^{-n}$ increases and is bounded above, so it converges to its supremum $1$ (monotone convergence). Watch the tail diameter shrink to $0$: every convergent sequence is Cauchy.
• Lim sup and Lim inf — Where the suprema and infima of tails settle: for $a_n = (-1)^n(1+1/n)$, $\limsup a_n = 1$ and $\liminf a_n = -1$. $\limsup a_n = \liminf a_n \iff \lim a_n$ exists.
• Stolz–Cesàro Theorem — The sequence version of l'Hôpital: if $(b_n)$ is strictly increasing and unbounded and $\lim \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = L$, then $\lim \frac{a_n}{b_n} = L$. Watch the difference ratio settle at $\frac{1}{3}$ first and the full ratio follow for $a_n = 1^2+\cdots+n^2$, $b_n = n^3$.