Vishvas's notes

Convergent sequences $(s_{n), (t_{n})$ in a field

Also see Topology ref, properties noted in lin alg ref.

Sums, products, quotients (sans 0 denominator) of convergent sequences are convergent.

Boundedness

Upper and lower bounds.

Montonicity

Monotone increasing and decreasing sequences.

A monotone increasing sequence that is bounded above must converge: imagine points on real line; similarly monotone decreasing sequence that is bounded below.

(Bolzano, Weierstrass): Every bounded sequence has a subsequence which converges: let M be the bound; either [-M, 0] or [0, M] has $\infty$ elements; so use this repeatedly to find monotone, bounded subsequence; this converges. Also holds for $R^{n}$ .

Upper and lower limits

Take set of subsequential limits E. Upper limit of $s_{n}$ : $s^{} = \limsup_{n \to \infty}s_{n} := \sup E$; similarly lower limit: $s_{} = \liminf_{n \to \infty}s_{n}$.

If $s_{n} \to s, s^{} = s_{} = s$. E is closed, so $s_{}, s^{} \in E$.

Generating function of a sequence

Take sequence $(a_{i})$ . Get generating function: $G (x; a_{n}) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ .

Series

$\infty$ series. A sequence of partial sums: $s_{n} = \sum_{i = 1}^{n} a_{i}$ .

Convergence

Many ideas from convergence of sequences.

Convergence tests

$s_{n}$ converges $\equiv$ it is cauchy. Aka completeness property of R or C. Cauchy criteria for convergence of $s_{n}$ : $\forall ϵ, \exists N : | s_{N} - s_{N + n} | < ϵ$ . So, $| a_{n} | \to 0$ .

Sum or product of absolutely convergent series is absolutely convergent: limit of new series is sum or product of limits (Cauchy product).

Comparison test: If $| a_{n} | \leq c_{n}$ : if $\sum c_{n}$ converges, $\sum a_{n}$ converges; if $\sum | a_{n} |$ diverges, $\sum c_{n}$ diverges: using Cauchy criterion.

If $a_{k} \geq a_{k + 1} \geq 0, s_{n} = \sum a_{n}$ converges $\equiv$ $t_{n} = \sum_{k = 0}^{n} 2^{k} a_{2^{k}}$ converges: $s_{n}$ monotonically increasing, so check boundedness: $\frac{t_{k}}{2} \geq s_{n} \leq t_{k}$ for $n \leq 2^{k}$ .

Ratio test

Converges if $lim sup | \frac{a_{n + 1}}{a_{n}} | < b < 1$ : bound by geometric series $b^{k} a_{n}$ ; diverges if $> 1 : | a_{n} | ↛ 0$ .

Root test

Take $x = lim sup | a_{n} |^{1 / n}$ : So $| a_{n} | \leq x^{n}$ . So, convergence if $x < 1$ , divergence if $x > 1$ ; No info otherwise: $\sum 1 / n ↗$ , but $\sum 1 / n^{2} \to c$ .

More powerful than ratio test: \ $lim inf | \frac{a_{n + 1}}{a_{n}} | \leq lim inf | a_{n} |^{1 / n} \leq lim sup | a_{n} |^{1 / n} \leq lim sup | \frac{a_{n + 1}}{a_{n}} |$ . \why

Alternating series test

If $| c_{i} | \geq | c_{i + 1} |; c_{2 k + 1} \geq 0, c_{2 k} \leq 0, lim c_{n} = 0$ , then $\sum c_{n}$ converges \why. So, $\frac{(- 1)^{n}}{n}$ converges.

Conditional convergence

Absolute ( $\sum | a_{i} |$ ) vs conditional ( $\sum a_{i}$ ) convergence. Absolute convergence $⟹$ conditional convergence: ratio test.

For conditional convergent series: There are $\infty$ +ve and -ve numbers which converge to $+ \infty$ and $- \infty$ : else contradicts absence of absolute convergence or presence of conditional convergence.

Absolute convergence

All $a_{n} \geq 0$ . $s_{n}$ monotone: so converges $\equiv$ it is bounded.

Rearrangement

Take $\sum a_{n}$ conditionally but not absolutely convergent: so $| a_{n} | \to 0$ , $x \leq y$ . $\exists$ rearrangement of terms: $s_{n}^{'} = \sum a_{k}^{'}$ , with $lim inf s_{n}^{'} = x, lim sup s_{n}^{'} = y$ : take +ve ${p_{n}}$ , -ve ${q_{n}}$ : both diverge, collect enough $p_{n}$ to go just over y; collect enough -ve numbers to go just under x; repeat: visualize as displacement between 2 points in R. So, can converge to any real number r. Eg: Rearrange terms of alternating harmonic series.

If $\sum a_{n}$ converges absolutely, all rearrangements converge to same limit: Any $s_{n}^{'} = \sum a_{k}^{'}$ is in some $s_{N}$ and vice versa.

Series product

$\sum a_{n}, \sum b_{n} : c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ is the product: put all ordered sequences $(a_{i}, b_{i})$ on a line: the diagonals of $a_{i} v s b_{i}$ matrix are $(c_{n})$ .

If $\sum | a_{n} |$ converges, $\sum a_{n} \to A, \sum b_{n} \to B, \sum c_{n} \to A B$ : Take $D_{n} = B_{n} - B$ , $C_{n} = \sum_{k = 0}^{n} a_{i} B_{n - i} = \sum_{k = 0}^{n} a_{i} (B + D_{n}) = A_{n} B + γ \to A B$ .

Special series

Arithmatic series

Arithmatic series: $S_{n} = \sum_{n} T n : T_{n} = a + n d$ . Reverse the series, add respective terms, get: $2 S_{n} = n (a + n d)$

Geometric-like series

$T_{n} = a r^{n}$ . Multiply by r, subtract terms, get: $(1 - r) S_{n} = a - a r^{n + 1}$ . Convergence as $n \to i n f t y$ : if $r < 1$ . Similar techniques for $a + 2 a r + 3 a r^{2} \dots$ .

Harmonic-like series

General harmonic series has: $T_{n} = (a + n d)^{- 1}$ .

$\sum 1 / n$ diverges: $1 + 1 / 2 + 1 / 3 + 1 / 4 . .$ : $\sum_{n = 3}^{4} 1 / n, \sum_{n = 5}^{9} 1 / n$ etc exceed 1/2. Harmonic number: $H (n) \approx \ln n = \int x^{- 1} d x$ .

$\sum 1 / n^{p}$ converges if $p > 1$ , diverges if $p \leq 1$ : check convergence of $\sum \frac{2^{k}}{2^{k p}}$ .

$\sum_{n = 2}^{\infty} \frac{1}{n (\log n)^{p}}$ converges $\equiv$ $p > 1$ .

Power series

$z \in C : \sum c_{n} z^{n}$ . If this converges, got Generating Function!

Take $a = \underset{n \to \infty}{lim sup} | c_{n} |^{1 / n}, R = 1 / a$ : convergence if $| z | < R$ (radius of convergence), divergence if $| z | > R$ : From Root test, ratio test.

If R = 1, $c_{i} \geq c_{i + 1} \geq 0$ , then $\sum c_{i} z^{n}$ converges $\forall z$ , except maybe z = 1. \why

Binomial theorem

For $(x + y)^{n} : n \in N$ : from combinatorics. For $(1 + x)^{n} : n \in R$ : Solve for coefficients in $(1 + x)^{n} = 1 + n x + . .$ by differentiating repeatedly.

e

$e = lim_{n \to \infty} (1 + n^{- 1})^{n} = lim_{n \to \infty} = 1 + 1 + \frac{1}{2!} . .$ . $2 < e < 1 + \sum_{n = 0}^{\infty} (\frac{1}{2})^{n} = 3$ , and e is an increasing sequence: so it converges.

Observe relationship with definition using natural log in another section.

Summation tricks

Summation by parts

Like integration by parts. Seq $(a_{n}), (b_{n}), A_{n} = \sum a_{n}$ . If $p \in [0, q], a_{n} = A_{n} - A_{n - 1}, \sum_{n = p}^{q} a_{n} b_{n} = \sum_{n = p}^{q - 1} A_{n} (b_{n} - b_{n + 1}) + A_{q} b_{q} - A_{p - 1} b_{p}$ .

If $(A_{n})$ bounded seq, $b_{0} \geq b_{1} . ., lim_{n \to \infty} b_{n} = 0, \sum b_{n} a_{n}$ converges: Can use summation by parts to show Cauchy criterion.

Using binomial summation formula

$\sum k = \sum (\binom{k}{1}) = (\binom{k + 1}{2})$ . Similarly, any polynomial $\sum a_{i} k^{i} = \sum b_{i} (\binom{k}{i})$ .

Derivative to estimate limit of partial sums

$(1 - x)^{- 1} = \sum_{i = 0}^{\infty} x^{i}$ ; take derivatives to get: $(1 - x)^{- 2} = \sum_{i = 0}^{\infty} i x^{i - 1} = \sum_{i = 0}^{\infty} (i - 1) x^{i - 1} - x^{i - 1}$ . Use similar trick to find: $i^{2} x^{i}$ .

Integral to estimate limit of partial sums

Also, use $\sum n \approx \int x d x$ .

The series $- \log (1 - x) = \int (\frac{1}{1 - x}) = \sum_{i = 1}^{\infty} x^{i} / i$ ; also from McLaurin series.

N to R

Convergent sequences \((s_{n), (t_{n})\) in a field