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    Sequences and series of numbers

    For a real sequence \((a_n)\), \(\limsup_na_n\) and \(\liminf_na_n\) exist in \([-\infty,\infty]\), and \((a_n)\) converges if and only if they are equal and finite. If \(\sum_na_n\) converges, then \(a_n\to0\) and the tails \(\sum_{n>N}a_n\) tend to \(0\). The comparison, root, and ratio tests, and for nonincreasing nonnegative terms the Cauchy condensation test, decide convergence by comparison with a series of known behaviour, and \(\sum_nn^{-p}\) converges if and only if \(p>1\).

    Limit superior and limit inferior of a sequence

    Let \((a_n)_{n\geq 0}\) be a sequence of real numbers. Its limit superior and limit inferior are \[ \begin{aligned} \limsup_{n\to\infty} a_n &\coloneqq \lim_{n\to \infty} \sup_{j\geq n} a_j = \inf_{n\geq 0} \sup_{j\geq n} a_j, \\ \liminf_{n\to\infty} a_n &\coloneqq \lim_{n\to \infty} \inf_{j\geq n} a_j = \sup_{n\geq 0} \inf_{j\geq n} a_j, \end{aligned} \] with values in \([-\infty,\infty]\); the limits exist and equal the infimum and supremum because \(\sup_{j\geq n}a_j\) is nonincreasing and \(\inf_{j\geq n}a_j\) is nondecreasing in \(n\).

    Convergent series have terms and tails tending to zero

    A convergent series has vanishing terms and vanishing tails.Suggest a slogan

    Let \((a_n)_{n\geq1}\) be a sequence in \(\CC\) such that \(\sum_{n=1}^\infty a_n\) converges. Then \(a_n\to0\) as \(n\to\infty\), and \[ \sum_{n=N}^\infty a_n \convergesto{N\to\infty} 0 . \]

    Geometric series

    The geometric series $\sum x^k$ converges exactly for $\abs{x}<1$, with sum $1/(1-x)$.Suggest a slogan

    Let \(x\in\CC\). The series \(\sum_{k=0}^\infty x^k\) converges if and only if \(\abs{x} < 1\), and in that case \[ \sum_{k=0}^\infty x^k = \frac 1 {1-x} . \]

    Comparison test for series with nonnegative terms

    If $0\le a_n\le b_n$, convergence of $\sum b_n$ implies convergence of $\sum a_n$, and divergence of $\sum a_n$ implies divergence of $\sum b_n$.Suggest a slogan

    Let \((a_n)_{n\geq1}\) and \((b_n)_{n\geq1}\) be real sequences with \(0\leq a_n \leq b_n\) for all \(n\).

    • If \(\sum_n b_n < \infty\), then \(\sum_n a_n < \infty\).

    • If \(\sum_n a_n = \infty\), then \(\sum_n b_n = \infty\).

    The Cauchy condensation test

    For $(a_k)$ nonincreasing and nonnegative, $\sum a_k$ converges exactly when $\sum 2^k a_{2^k}$ converges.Suggest a slogan

    Let \((a_k)_{k\geq1}\) be a nonincreasing sequence of nonnegative real numbers. Then \[ \sum_{k\geq 1} a_k < \infty \iff \sum_{k\geq 1} 2^k a_{2^k}<\infty . \]

    Group the terms in dyadic blocks \(2^k\le j<2^{k+1}\). Since \((a_j)\) is nonincreasing, each block satisfies \[ 2^k a_{2^{k+1}}\le\sum_{j=2^k}^{2^{k+1}-1}a_j\le 2^k a_{2^k}. \] Summing over \(0\le k\le K\) gives \[ \frac12\sum_{k=1}^{K+1}2^{k}a_{2^{k}}\le\sum_{j=1}^{2^{K+1}-1}a_j\le\sum_{k=0}^{K}2^k a_{2^k}. \] Both series have nonnegative terms, so their partial sums are nondecreasing, and these bounds show that one sequence of partial sums is bounded if and only if the other is; the term \(a_1\) omitted from \(\sum_{k\geq1}2^ka_{2^k}\) does not affect convergence.

    $p$-tests for series and integrals

    $\sum k^{-p}$ and $\int_\varepsilon^\infty x^{-p}\,dx$ converge exactly for $p>1$ and $\int_0^1 x^{-p}\,dx$ exactly for $p<1$; in $\RR^n$ the threshold for $\abs{x}^{-p}$ is $p=n$.Suggest a slogan

    Let \(p\in\RR\), let \(\varepsilon>0\), let \(n\geq1\), and let \(B \coloneqq \theset{x\in \RR^n \suchthat \abs{x} \leq 1}\) be the closed unit ball for the Euclidean norm. Then \[ \begin{aligned} \sum_{k=1}^\infty \frac 1 {k^p} < \infty &\iff p>1, \\ \int_\varepsilon^\infty \frac {dx} {x^p} < \infty &\iff p>1, \\ \int_0^1 \frac {dx} {x^p} < \infty &\iff p<1, \\ \int_B \frac{dx}{\abs{x}^p} < \infty &\iff p < n, \\ \int_{\RR^n\setminus B} \frac{dx}{\abs{x}^p} < \infty &\iff p > n . \end{aligned} \]

    Bernoulli's inequality

    For $r\ge1$ and $x\ge-1$, convexity gives $(1+x)^r\ge1+rx$.Suggest a slogan

    Let \(x\in\RR\).

    1. If \(n\geq 0\) is an integer and \(x\geq -1\), then \((1+x)^n \geq 1 + nx\).

    2. If \(n\geq 0\) is an even integer, then \((1+x)^n \geq 1 + nx\) for every \(x\in\RR\).

    3. If \(r\geq 1\) is real and \(x\geq -1\), then \((1+x)^r \geq 1 + rx\).

    1. By induction on \(n\): the case \(n = 0\) is \(1\geq 1\), and if \((1+x)^n\geq 1+nx\), then since \(1+x\geq 0\), \[ (1+x)^{n+1}\geq(1+nx)(1+x) = 1 + (n+1)x + nx^2\geq 1+(n+1)x. \]

    2. For \(x\geq -1\) this is (1). For \(x<-1\) and even \(n\geq 2\), \((1+x)^n\geq 0\) while \(1+nx<1-n<0\); the case \(n=0\) is \(1\geq 1\).

    3. The function \(\varphi(t)\coloneqq(1+t)^r\) is convex on \([-1,\infty)\) because \(\varphi''(t) = r(r-1)(1+t)^{r-2}\geq 0\) for \(t>-1\) and \(\varphi\) is continuous at \(-1\), so \(\varphi\) lies above its tangent line \(1+rt\) at \(t = 0\).

    For all \(x\in\RR\), \[ 1-x \leq e^{-x} . \]

    State Bernoulli's inequality and say for which $x$ and $n$ it holds.