Theorems and proofs

Limit comparison

\begin{matrix} L = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | \\ \begin{matrix} 1. & L < 1 ⟹ a_{n} \to 0 \\ 2. & L > 1 ⟹ a_{n} \to \infty \\ 3. & L = 1 or ∄ L ⟹ Inconclusive \\ 4. & \sqrt[n]{| a_{n} |} \to L \end{matrix} \end{matrix}

Bernoulli

\begin{matrix} x > - 1 ⟹ (1 + x)^{n} \geq 1 + n x \end{matrix}

Bolzano-Weierstrass theorem

\begin{matrix} Every bounded sequence has a convergent subsequence \\ Proof: \\ \exists a_{k_{n}} : a_{k_{n}} monotonic and bounded \\ ⟹ a_{k_{n}} converges \end{matrix}

Cauchy's criterion

\begin{matrix} {a_{n}} \subseteq R \\ a_{n} converges ⟺ a_{n} is a Cauchy’s sequence \\ Proof: \\ Let a_{n} \to L \\ \forall ε > 0 \exists N_{ε} : \forall n > N_{ε} : | a_{n} - L | < \frac{ε}{2} \\ Let m \geq n \\ | a_{n} - a_{m} | = | a_{n} - L - a_{m} + L | \leq \underset{< \frac{ε}{2}}{\underset{⏟}{| a_{n} - L |}} + \underset{= | a_{m} - L | < \frac{ε}{2}}{\underset{⏟}{| L - a_{m} |}} \\ ⟹ | a_{n} - a_{m} | < ε ⟺ a_{n} is a Cauchy’s sequence \end{matrix}

\begin{matrix} Let a_{n} be a Cauchy’s sequence \\ 1. & Cauchy’s sequence is bounded \\ Proof: \\ By definition of the limit: \exists ε = 7 : \exists N_{ε} : \forall n, m > N_{ε} : | a_{n} - a_{m} | < 7 \\ Let z = a_{⌈ N_{ε} ⌉ + 1} \\ Then - 7 < a_{n} - z < 7 ⟺ \forall n > N_{ε} : z - 7 \leq a_{n} \leq z + 7 \\ A = {a_{1}, a_{2}, a_{3}, \dots, a_{⌈ N_{ε} ⌉}} is a finite set ⟹ \exists m = m i n (A), M = m a x (A) \\ ⟹ \forall n \in N : m i n (m, z - 7) \leq a_{n} \leq m a x (M, z + 7) - sequence is bounded \\ 2. & If sequence is bounded, then there exists a convergent subsequence a_{k_{n}} \\ \exists a_{k_{n}} \to L_{k} \\ 3. & Limit of a subsequence is a limit of the whole sequence, if such exists \\ \forall ε > 0 \exists N_{a} : \forall n, m > N_{a} : | a_{n} - a_{m} | < \frac{ε}{2} \\ \forall ε > 0 \exists N_{k} : \forall n > N_{k} : | a_{k_{n}} - L_{k} | < \frac{ε}{2} \\ Let n > N_{ε} = m a x (N_{a}, N_{k}) \\ | a_{n} - L_{k} | = | a_{n} - a_{k_{n}} - L_{k} + a_{k_{n}} | \leq \underset{< \frac{ε}{2}}{\underset{⏟}{| a_{n} - a_{k_{n}} |}} + \underset{< \frac{ε}{2}}{\underset{⏟}{| a_{k_{n}} - L_{k} |}} < ε \\ ⟺ a_{n} \to L_{k} \\ 4. & Proof for 3. via lim^{―} a_{n} \\ Let lim^{―} a_{n} = T > L_{k} \\ Then \exists a_{m_{n}} \to T \\ \exists N_{1} : \forall n > N_{1} : | a_{k_{n}} - L_{k} | < \frac{T - L}{4} \\ \exists N_{2} : \forall n > N_{2} : | a_{m_{n}} - T | < \frac{T - L}{4} \\ \exists N_{3} : \forall n, m > N_{3} : | a_{n} - a_{m} | < \frac{T - L}{4} \\ \exists N_{ε} = m a x (N_{1}, N_{2}, N_{3}) : \forall n, m > N_{ε} : \\ | a_{k_{n}} - L_{k} | < \frac{T - L}{4}, | a_{m_{n}} - T | < \frac{T - L}{4}, | a_{n} - a_{m} | < \frac{T - L}{4} \\ L - \frac{T - L}{4} < a_{k_{n}} < L + \frac{T - L}{4} \\ T - \frac{T - L}{4} < a_{m_{n}} < T + \frac{T - L}{4} \\ \frac{5 L - T}{4} < a_{k_{n}} < \frac{T + 3 L}{4} \\ \frac{3 T + L}{4} < a_{m_{n}} < \frac{5 T - L}{4} \\ a_{m_{n}} - a_{k_{n}} > \frac{3 T + L}{4} - a_{k_{n}} > \frac{3 T + L}{4} - \frac{T + 3 L}{4} = \frac{T - L}{2} > \frac{T - L}{4} \end{matrix}

e

{(1 + \frac{a}{n})}^{n} \to e^{a}

Harmonic series

\sum_{n = 1}^{\infty} \frac{1}{n^{a}} \to L ⟺ a > 1

Convergence of series

\begin{matrix} \sum_{n = 1}^{\infty} a_{n} \to L ⟺ S_{N} is a Cauchy’s sequence \\ (\forall ε > 0 : \exists N_{ε} : \forall N, M > N_{ε} : | S_{N} - S_{M} | < ε) \\ If M > N : \\ S_{M} = S_{N} + \sum_{n = N + 1}^{M} a_{n} \\ | S_{N} - S_{M} | < ε \\ | \sum_{n = N + 1}^{M} a_{n} | < ε \end{matrix}

Direct comparison test

\begin{matrix} \sum a_{n}, \sum b_{n} : \\ 0 \leq a_{n} \leq b_{n} \\ \sum b_{n} \to L ⟹ \sum a_{n} \to M \\ \sum a_{n} ↛ M ⟹ \sum b_{n} ↛ L \\ Proof: \\ 0 \leq a_{n} ⟹ A_{N} - A_{N - 1} = a_{N} \geq 0 \\ A_{N} > A_{N} ⟹ A_{N} is monotonically non-descending \\ \sum_{n = 1}^{N} b_{n} = B_{N} \\ \sum b_{n} \to L ⟹ \exists C : B_{N} \leq C \\ A_{N} \leq B_{N} \leq C ⟹ A_{N} < C \\ ⟹ A_{N} is monotonically non-descending and upper-boundeed ⟹ A_{N} \to M \end{matrix}

Limit comparison test

\begin{matrix} \sum a_{n}, \sum b_{n} \\ 0 \leq a_{n}, b_{n} \\ \exists lim_{n \to \infty} \frac{a_{n}}{b_{n}} = L \\ L > 0 ⟹ (\sum a_{n} \to M_{1} ⟺ \sum b_{n} \to M_{2}) \\ L = 0 ⟹ (\sum a_{n} \to M_{1} ⟸ \sum b_{n} \to M_{2}) \\ L = \infty ⟹ (\sum a_{n} \to M_{1} ⟹ \sum b_{n} \to M_{2}) \\ Proof: \\ Let 0 \leq a_{n}, b_{n} \\ Let lim_{n \to \infty} \frac{a_{n}}{b_{n}} = L \end{matrix}

\begin{matrix} Proof for 1. \\ L > 0 \\ ⟹ \frac{1}{e} L < \frac{a_{n}}{b_{n}} < e L \\ \frac{1}{e} L b_{n} \leq a_{n} \leq e L b_{n} \\ 1. \sum a_{n} \to M_{1} ⟹ \sum \frac{1}{e} b_{n} \to M^{'} ⟹ \sum b_{n} \to M_{2} \\ 2. \sum b_{n} \to M_{2} ⟹ \sum e L b_{n} \to M^{'} ⟹ \sum a_{n} \to M_{1} \\ 1. and 2. ⟹ \sum a_{n} \to M_{1} ⟺ \sum b_{n} \to M_{2} \end{matrix}

\begin{matrix} Proof for 2. \\ L = 0, \sum b_{n} \to M_{2} \\ \frac{a_{n}}{b_{n}} \to 0 ⟹ \frac{a_{n}}{b_{n}} < 7 ⟹ a_{n} < 7 b_{n} \\ \sum b_{n} \to M_{2} ⟹ \sum 7 b_{n} \to 7 M_{2} ⟹ \sum a_{n} \to M_{1} \end{matrix}

\begin{matrix} Proof for 3. \\ L = \infty, \sum a_{n} \to M_{1} \\ \frac{a_{n}}{b_{n}} \to \infty ⟹ \frac{a_{n}}{b_{n}} > 420 ⟹ a_{n} > 420 b_{n} \\ \sum a_{n} \to M_{1} ⟹ \sum 420 b_{n} \to M^{'} ⟹ \sum b_{n} \to M_{2} \end{matrix}

Root test (nth root test, Cauchy's criterion)

\begin{matrix} Let L = lim_{n \to \infty} \sqrt[n]{| a_{n} |} \\ L > 1 ⟹ \sum a_{n} ↛ M \\ L < 1 ⟹ \sum | a_{n} | \to M \\ L = 1 ⟹ Test is inconclusive \\ Proof: \\ Let \overset{―}{lim_{n \to \infty}} \sqrt[n]{| a_{n} |} \\ ⟹ \exists a_{k_{n}} : \sqrt[k_{n}]{| a_{k_{n}} |} \to L \\ If L > 1 ⟹ L > \frac{L + 1}{2} > 1 \\ ⟹ \sqrt[k_{n}]{| a_{k_{n}} |} \geq \frac{L + 1}{2} ⟺ | a_{k_{n}} | \geq {(\frac{L + 1}{2})}^{k_{n}} \\ \frac{L + 1}{2} > 1 ⟹ {(\frac{L + 1}{2})}^{k_{n}} \to \infty ⟹ | a_{k_{n}} | \to \infty \\ ⟹ a_{k_{n}} ↛ 0 ⟹ a_{n} ↛ 0 ⟹ \sum a_{n} ↛ M \\ If L < 1 ⟹ L < \frac{L + 1}{2} \\ \sqrt[k_{n}]{| a_{k_{n}} |} \to L ⟹ \sqrt[n]{| a_{n} |} \leq \frac{L + 1}{2} ⟹ | a_{n} | \leq {(\frac{L + 1}{2})}^{n} \\ \sum {(\frac{L + 1}{2})}^{n} \to M^{'} ⟹ \sum | a_{n} | \to M \end{matrix}

Ratio test (d'Alembert's criterion)

\begin{matrix} Let L = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | \\ L > 1 ⟹ \sum a_{n} ↛ M \\ L < 1 ⟹ \sum | a_{n} | \to M \end{matrix}

Cauchy's condensation test

\begin{matrix} Let a_{n} \geq 0, a_{n} is monotonically non-increasing \\ \sum a_{n} \to M_{1} ⟺ \sum_{n = 0}^{\infty} 2^{n} a_{2^{n}} \to M_{2} \\ Corollary: \\ \sum \frac{1}{n^{p}} \to M_{1} ⟺ p > 1 \\ \sum \frac{1}{n^{p}} \to M_{1} ⟺ \sum 2^{n} \frac{1}{(2^{n})^{p}} \to M_{2} \\ \sum 2^{n} \frac{1}{(2^{n})^{p}} = \sum 2^{n - n p} = \sum (2^{1 - p})^{n} \\ ⟹ \sum \frac{1}{n^{p}} \to M_{1} ⟺ 2^{1 - p} < 1 ⟺ 1 - p < 0 ⟺ p > 1 \end{matrix}

Dirichlet's test

\begin{matrix} \sum b_{n} is bounded \\ a_{n} monotonically non-increasing, a_{n} \to 0 \\ Then \sum b_{n} a_{n} \to M_{1} \end{matrix}

\begin{matrix} Example: \\ \sum \frac{(- 1)^{n}}{n} = \sum (- 1)^{n} \cdot \frac{1}{n} \\ | \sum (- 1)^{n} | \leq 1 \\ \frac{1}{n} \to 0 \\ ⟹ \sum \frac{(- 1)^{n}}{n} \to M_{1} \end{matrix}

\begin{matrix} Proof: \\ a_{n} \to 0 and a_{n} is monotonically non-increasing \\ ⟹ \forall n \in N : a_{n} \geq 0 \\ \sum b_{n} is bounded ⟹ S_{N} = \sum_{n = 1}^{N} b_{n} is bounded \\ \sum a_{n} b_{n} \to L ⟺ \forall ε > 0 : \exists N_{ε} : \forall N, M > N_{ε} \in N : | \sum_{n = M + 1}^{N} a_{n} b_{n} | < ε \\ | \sum_{n = M + 1}^{N} a_{n} b_{n} | = | \sum_{n = M + 1}^{N} a_{n} (S_{n} - S_{n - 1}) | = | \sum_{n = M + 1}^{N} a_{n} S_{n} - \sum_{n = M + 1}^{N} a_{n} S_{n - 1} | = \\ = | - S_{M} a_{M + 1} + \sum_{n = M + 1}^{N - 1} S_{n} (a_{n} - a_{n + 1}) + S_{N} a_{N} | \leq \\ \leq | - S_{M} a_{M + 1} | + \sum_{n = M + 1}^{N - 1} | S_{n} (a_{n} - a_{n + 1}) | + | S_{N} a_{N} | = \\ = | S_{M} | \cdot | a_{M + 1} | + \sum_{n = M + 1}^{N - 1} | S_{n} | \cdot | (a_{n} - a_{n + 1}) | + | S_{N} | \cdot | a_{N} | \\ | S_{n} | \leq K \\ a_{N}, a_{M + 1} \geq 0 \\ a_{M + 1} \geq a_{M + 2} \geq \dots \\ ⟹ a_{M + 1} - a_{M + 2} \geq 0 \\ ⟹ | S_{M} | \cdot | a_{M + 1} | + \sum_{n = M + 1}^{N - 1} | S_{n} | \cdot | (a_{n} - a_{n + 1}) | + | S_{N} | \cdot | a_{N} | \leq \\ K \cdot (a_{M + 1} + \sum_{n = M + 1}^{N} (a_{n} - a_{n + 1}) + a_{N}) \\ ⟹ | \sum_{n = M + 1}^{N} a_{n} b_{n} | \leq K \cdot (a_{M + 1} + a_{M + 1} - a_{M + 2} + \dots + a_{N - 1} - a_{N} + a_{N}) = \\ = K \cdot 2 a_{M + 1} \\ | \sum_{n = M + 1}^{N} a_{n} b_{n} | \leq 2 K \cdot a_{M + 1} \\ \forall ε > 0 : \exists N_{ε} : \forall M > N_{ε} : | a_{M + 1} | < \frac{ε}{2 K} \\ ⟹ \forall ε > 0 : \exists N_{ε} : \forall M > N_{ε} : | \sum_{n = M + 1}^{N} a_{n} b_{n} | \leq 2 K \cdot a_{m + 1} < 2 K \cdot \frac{ε}{2 K} = ε \end{matrix}

\begin{matrix} \sum (- 1)^{n}, \sum \sin (n), \sum \cos (n) are bounded series \end{matrix}

\begin{matrix} \sum \sin (n) is bounded ⟹ \sum \frac{\sin (n)}{n} = \sum (\sin (n) \cdot \underset{\to 0}{\underset{⏟}{\frac{1}{n}}}) \to M_{1} \end{matrix}

\begin{matrix} S_{N} = \sum_{n = 1}^{N} \sin (n) = \sin (1) + \sin (2) + \dots + \sin (N) \\ 2 \sin (x) \sin (y) = \cos (x - y) - \cos (x + y) \\ 2 \sin (1) S_{N} = 2 \sin (1) \sin (1) + 2 \sin (1) \sin (2) + \dots + 2 \sin (1) \sin (N) = \\ = \cos (0) - \cos (2) + \cos (1) - \cos (3) + \cos (2) - \cos (4) + \dots + \cos (N - 1) - \cos (N + 1) = \\ = \cos (0) + \cos (1) - \cos (N) - \cos (N + 1) \\ ⟹ S_{N} = \frac{\cos (0) + \cos (1) - \cos (N) - \cos (N + 1)}{2 \sin (1)} \\ - 2 \leq - \cos (N) - \cos (N + 1) \leq 2 ⟹ \frac{\cos (0) + \cos (1) - 2}{2 \sin (1)} \leq S_{N} \leq \frac{\cos (0) + \cos (1) + 2}{2 \sin (1)} \end{matrix}

Alternating series test (Leibniz criterion)

\begin{matrix} \sum (- 1)^{n} \\ a_{n} is monotonically non-increasing and a_{n} \to 0 \\ \sum (- 1)^{n} a_{n} \to M_{1} \end{matrix}

Inverse derivative

\begin{matrix} Let f : [a, b] \to [c, d] be continuous and invertible \\ Let y \in (c, d) \\ Let f be differentiable at f^{- 1} (y) and f^{'} (f^{- 1} (y)) \neq 0 \\ Then f^{- 1} is differentiable at y and (f^{- 1})^{'} (y) = \frac{1}{f^{'} (f^{- 1} (y))} \\ Proof: \\ (f^{- 1})^{'} (y) = lim_{t \to y} \frac{f^{- 1} (t) - f^{- 1} (y)}{t - y} \\ Let x = f^{- 1} (y), z = f^{- 1} (t) \\ y = f (x), t = f (z) \\ t \to y ⟹ f (z) \to f (x) \\ ⟹ f^{- 1} (f (z)) \to f^{- 1} (f (x)) ⟹ z \to x \\ (f^{- 1})^{'} (y) = lim_{t \to y} \frac{f^{- 1} (t) - f^{- 1} (y)}{t - y} = \\ = lim_{z \to x} \frac{z - x}{f (z) - f (x)} = lim_{z \to x} \frac{1}{\frac{f (z) - f (x)}{z - x}} \\ f^{'} (x) = f^{'} (f^{- 1} (y)) = lim_{z \to x} \frac{f (z) - f (x)}{z - x} \neq 0 \\ ⟹ (f^{- 1})^{'} (y) = lim_{z \to x} \frac{1}{\frac{f (z) - f (x)}{z - x}} = \frac{1}{f^{'} (x)} = \frac{1}{f^{'} (f^{- 1} (y))} \end{matrix}

Intermediate value theorem

\begin{matrix} Let f be a continuous function on [a, b] \\ Then f (a) \cdot f (b) < 0 ⟹ \exists x \in [a, b] : f (x) = 0 \end{matrix}

Weierstrass boundedness theorem

\begin{matrix} Let f be a continuous function on [a, b] \\ Then \exists m, M : \forall x \in [a, b] : m \leq f (x) \leq M \\ Proof: \\ Let f be unbounded from above on [a, b] \\ \forall n \in N : \exists x_{n} : f (x_{n}) > n \\ [a, b] is bounded ⟹ By Bolzano-Weierstrass theorem \exists x_{n_{k}} \to x : x \in [a, b] \\ f is continuous on [a, b] ⟹ f (x_{n_{k}}) \to f (x) \in R \\ f (x_{n}) \to \infty ⟹ f (x_{n_{k}}) \to \infty - Contradiction! \\ Similar proof for bounded from below \\ ⟹ f is bounded on [a, b] \end{matrix}

Weierstrass extreme value theorem

\begin{matrix} Let f be a continuous function on [a, b] \\ Then \exists c, d \in [a, b] : \forall x \in [a, b] : f (c) \leq f (x) \leq f (d) \\ Proof: \\ f is bounded from above ⟹ \exists M = s u p (f (x)) on [a, b] \\ Let n \in N \\ M - \frac{1}{n} is not an upper bound of f on [a, b] \\ ⟹ \exists d_{n} \in [a, b] : M - \frac{1}{n} < f (d_{n}) \\ \forall n \in N : M - \frac{1}{n} < f (d_{n}) \leq M \\ ⟹ f (d_{n}) \to M \\ By Bolzano-Weierstrass theorem: \exists d_{n_{k}} \to d \in [a, b] \\ f is continuous ⟹ f (d_{n_{k}}) \to f (d) \\ ⟹ f (d_{n}) \to f (d) ⟹ M = f (d) \\ Similar proof for minimum \end{matrix}

Fermat's theorem (stationary points)

\begin{matrix} Let f be defined on (a, b) \\ Let x_{0} \in (a, b) \\ If f has a local extremum at x_{0} and f is differentiable at x_{0} \\ Then f^{'} (x_{0}) = 0 \\ Proof: \\ Let x be a local minimum \\ f^{'} (x_{0}) = lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}} \\ lim_{x \to x_{0}^{+}} \frac{f (x) - f (x_{0})}{x - x_{0}} \geq 0 \\ lim_{x \to x_{0}^{-}} \frac{f (x) - f (x_{0})}{x - x_{0}} \leq 0 \\ lim_{x \to x_{0}^{+}} \frac{f (x) - f (x_{0})}{x - x_{0}} = lim_{x \to x_{0}^{-}} \frac{f (x) - f (x_{0})}{x - x_{0}} \\ ⟹ f^{'} (x_{0}) = 0 \\ Similar proof for local maximum \end{matrix}

Rolle's theorem

\begin{matrix} Let f be continuous on [a, b] and differentiable on (a, b) \\ Let f (a) = f (b) \\ Then \exists c \in (a, b) : f^{'} (c) = 0 \\ Proof: \\ f is continuous at [a, b] ⟹ \exists c, d \in [a, b] : \forall x \in [a, b] : f (c) \leq f (x) \leq f (d) \\ \dots \end{matrix}

Darboux's theorem

\begin{matrix} Let f be a differentiable function on [a, b] \\ Then \forall a_{1} < b_{1} \in [a, b] : \forall y \in [f^{'} (a_{1}), f^{'} (b_{1})] \exists x \in [a, b] : f^{'} (x) = y \\ In other words, f^{'} has intermediate value property \\ Note: f^{'} does not have to be differentiable on [a, b] \\ If it is not continuous, then the discontinuities will be essential \end{matrix}

L'Hopital's theorem/rule

\begin{matrix} Let f, g be differentiable functions in the locality of a \\ Let \exists lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} \\ Let lim_{x \to a} f (x) = lim_{x \to a} g (x) = 0 or \pm \infty \\ Then lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} \end{matrix}

Lagrange's mean value theorem

\begin{matrix} This is a generalization of Rolle’s theorem \\ Let f be continuous on [a, b] and differentiable on (a, b) \\ Then \exists ξ \in (a, b) : f^{'} (ξ) = \frac{f (b) - f (a)}{b - a} \end{matrix}

Cauchy's theorem

\begin{matrix} Let f, g be continuous on [a, b] \\ and differentiable on (a, b) \\ Let \forall x \in (a, b) : g^{'} (x) \neq 0 \\ Then \exists c \in (a, b) : \frac{f^{'} (c)}{g^{'} (c)} = \frac{f (b) - f (a)}{g (b) - g (a)} \end{matrix}