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 \(s_{n}\): \(\forall \eps, \exists N: |s_{N}-s_{N+n}|< \eps\). 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 \(\limsup |\frac{a_{n+1}}{a_{n}}| < b< 1\): bound by geometric series \(b^{k}a_{n}\); diverges if \(>1: |a_{n}| \notto 0\).
Root test
Take \(x = \limsup |a_{n}|^{1/n}\): So \(|a_{n}| \leq x^{n}\). So, convergence if \(x<1\), divergence if \(x>1\); No info otherwise: \(\sum 1/n \diverge\), but \(\sum 1/n^{2} \to c\).
More powerful than ratio test: \ \(\liminf |\frac{a_{n+1}}{a_{n}}| \leq \liminf |a_{n}|^{1/n} \leq \limsup |a_{n}|^{1/n} \leq \limsup |\frac{a_{n+1}}{a_{n}}|\). \why
Alternating series test
If \(|c_{i}| \geq |c_{i+1}|; c_{2k+1} \geq 0, c_{2k} \leq 0, \lim c_{n} = 0\), 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 \(\implies\) 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 \(\liminf s_{n}’ = x, \limsup s_{n}’ = y\): take +ve \(\set{p_{n}}\), -ve \(\set{q_n}\): 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 \(a_{i} vs b_{i}\) matrix are \((c_{n})\).
If \(\sum |a_{n}|\) converges, \(\sum a_{n} \to A, \sum b_{n} \to B, \sum c_{n} \to AB\): Take \(D_n = B_n - B\), \(C_{n} = \sum_{k=0}^{n}a_{i}B_{n-i} = \sum_{k=0}^{n}a_{i}(B + D_n) = A_{n}B + \gamma \to AB\).
Special series
Arithmatic series
Arithmatic series: \(S_n = \sum_n Tn: T_n = a + nd\). Reverse the series, add respective terms, get: \(2S_n = n(a+nd)\)
Geometric-like series
\(T_n = ar^{n}\). Multiply by r, subtract terms, get: \((1 - r)S_n = a - ar^{n+1}\). Convergence as \(n \to infty\): if \(r < 1\). Similar techniques for \(a+2ar+3ar^{2} \dots\).
Harmonic-like series
General harmonic series has: \(T_n = (a + nd)^{-1}\).
\(\sum 1/n \) 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: \(H(n)\approx \ln n = \int x^{-1} dx\).
\(\sum 1/n^{p}\) converges if \(p>1\), diverges if \(p \leq 1\): check convergence of \(\sum \frac{2^{k}}{2^{kp}}\).
\(\sum_{n=2}^{\infty} \frac{1}{n (\log n)^{p}}\) converges \(\equiv\) \(p>1\).
Power series
\(z\in C: \sum c_{n}z^{n}\). If this converges, got Generating Function!
Take \(a = \limsup_{n \to \infty} |c_{n}|^{1/n}, R = 1/a\): 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 \((1+x)^{n} = 1 + nx + .. \) 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: \((1-x)^{-2} = \sum_{i=0}^{\infty} ix^{i-1} = \sum_{i=0}^{\infty} (i-1)x^{i-1} - x^{i-1}\). Use similar trick to find: \(i^{2}x^{i}\).
Integral to estimate limit of partial sums
Also, use \(\sum n \approx \int x dx\).
The series \(-\log (1 - x) = \int(\frac{1}{1-x}) = \sum_{i=1}^{\infty} x^{i}/i\); also from McLaurin series.