Proof of Standard Examples of Series: the Harmonic Series, the Reciprocals of the Squares, a Centred Quadratic Sum and the Exponential Series

Each result cited is universally quantified over the data in its own statement.

Preliminaries. A sum $\sum_{k=1}^{n}a_{k}$ with $n\in\mathbb{N}$ is the sum over the set $[n]$ and agrees with the iterated sum of Iterated Operations: Finite Sums and Finite Products §iterated, by Sums and Products over a Finite Set and over an Interval §intervals; so Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms applies to it. Sums, products, differences and inequalities of natural numbers are the same in $\mathbb{N}_{0}$ and in $\mathbb{R}$, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement. For $n\in\mathbb{N}$ one has $1\le n$, so $n$ is a positive real number; and for $k\in\mathbb{N}_{0}$, $1\le k!$ by The Factorial: Recursion, Positivity and Bounds by Powers §positive, so $k!$ is positive. For $y\in\mathbb{R}$, $y^{2}=y\cdot y$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product, as $2=1+1$; hence $y^{2}\ge0$, $(-y)^{2}=y^{2}$, $n^{2}>0$ for $n\in\mathbb{N}$, and $1+\alpha y^{2}\ge1>0$. So every denominator in the statement is positive.

Clause harmonic. For $n\in\mathbb{N}$ let $H_{n}=\sum_{m=1}^{n}\frac{1}{m}$; these are the partial sums of $\sum_{m=1}^{\infty}\frac{1}{m}$. Let $n\in\mathbb{N}$. By Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §splitting, $H_{n+n}=H_{n}+\sum_{j=1}^{n}\frac{1}{n+j}$. For $j\in[n]$ one has $j\le n$, hence $n+j\le n+n$, so $0<n+j\le n+n$ in $\mathbb{R}$ and $\frac{1}{n+n}\le\frac{1}{n+j}$. By Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §comparison and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §constant, $\sum_{j=1}^{n}\frac{1}{n+j}\ge n\cdot\frac{1}{n+n}=\frac{1}{2}$, so

$$H_{n+n}\ge H_{n}+\frac{1}{2}\qquad(n\in\mathbb{N}).\tag{1}$$

Let $A$ be the class of those $j\in\mathbb{N}$ for which there is $n\in\mathbb{N}$ with $H_{n}\ge\frac{j}{2}$. Then $1\in A$, witnessed by $n=1$, since $H_{1}=\frac{1}{1}=1\ge\frac{1}{2}$ by Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion. If $j\in A$, witnessed by $n$, then (1) gives $H_{n+n}\ge\frac{j}{2}+\frac{1}{2}=\frac{j+1}{2}$, so $j+1\in A$. By The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction, $A=\mathbb{N}$. Suppose now that $b\in\mathbb{R}$ were an upper bound of $\{H_{n}:n\in\mathbb{N}\}$. Choose first, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §archimedean applied to $2b$, some $j\in\mathbb{N}$ with $2b<j$; then, as $j\in A$, choose $n\in\mathbb{N}$ with $H_{n}\ge\frac{j}{2}$. Then $b<\frac{j}{2}\le H_{n}\le b$, which is impossible. So the set has no upper bound, that is, it is not bounded above in the sense of Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded, and by Bounded Sequences of Real Numbers §bounded the sequence $(H_{n})$ is not bounded above. The terms $\frac{1}{m}$ are positive, so by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series §bounded the series $\sum_{m=1}^{\infty}\frac{1}{m}$ does not converge, that is, it diverges.

Clause reciprocal-squares. Let $P$ be the class of those $N\in\mathbb{N}$ with $\sum_{m=1}^{N}\frac{1}{m^{2}}\le2-\frac{1}{N}$. By Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion and $1^{2}=1$ (Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product), $\sum_{m=1}^{1}\frac{1}{m^{2}}=1=2-\frac{1}{1}$, so $1\in P$. Let $N\in P$. By Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion,

$$\sum_{m=1}^{N+1}\frac{1}{m^{2}}=\sum_{m=1}^{N}\frac{1}{m^{2}}+\frac{1}{(N+1)^{2}}.$$

From $0<N\le N+1$ and $0<N+1$ we get $0<N(N+1)\le(N+1)(N+1)=(N+1)^{2}$, hence $\frac{1}{(N+1)^{2}}\le\frac{1}{N(N+1)}=\frac{1}{N}-\frac{1}{N+1}$. Adding this to the inequality for $N$ gives $\sum_{m=1}^{N+1}\frac{1}{m^{2}}\le2-\frac{1}{N}+\frac{1}{N}-\frac{1}{N+1}=2-\frac{1}{N+1}$, so $N+1\in P$. By The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction, $P=\mathbb{N}$; in particular $M\in P$. The terms $\frac{1}{m^{2}}$ are positive, and the partial sums $s_{N}=\sum_{m=1}^{N}\frac{1}{m^{2}}$ satisfy $s_{N}\le2-\frac{1}{N}\le2$ for every $N\in\mathbb{N}$, so $(s_{N})$ is bounded above by Bounded Sequences of Real Numbers §bounded and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded. By Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series §bounded the series $\sum_{m=1}^{\infty}\frac{1}{m^{2}}$ converges with sum $\sup\{s_{N}:N\in\mathbb{N}\}$, which is at most $2$ because $2$ is an upper bound of that set and the supremum is its least upper bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum).

Clause centred. For $m\in[2M+1]$ put $c_{m}=\frac{1}{1+\alpha(m-M-1)^{2}}$, and for $r\in[M]$ put $e_{r}=\frac{1}{1+\alpha r^{2}}$. Since $2M+1=M+(1+M)$, two applications of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §splitting and one of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §recursion give

$$\sum_{m=1}^{2M+1}c_{m}=\sum_{m=1}^{M}c_{m}+\sum_{j=1}^{1+M}c_{M+j}=\sum_{m=1}^{M}c_{m}+c_{M+1}+\sum_{r=1}^{M}c_{M+1+r}.$$

In $\mathbb{R}$, $(M+1)-M-1=0$, so $c_{M+1}=1$; and $(M+1+r)-M-1=r$, so $c_{M+1+r}=e_{r}$ for $r\in[M]$. By Iterated Operations over an Interval: Extracting One Term and Reversing the Order §reversal, applied to the restriction of $c$ to $[M]$, $\sum_{m=1}^{M}c_{m}=\sum_{r=1}^{M}c_{M+1-r}$; and in $\mathbb{R}$, $(M+1-r)-M-1=-r$, so $c_{M+1-r}=e_{r}$ by $(-r)^{2}=r^{2}$. Hence

$$\sum_{m=1}^{2M+1}c_{m}=1+2\sum_{r=1}^{M}e_{r}.$$

For $r\in[M]$, $0<\alpha r^{2}<1+\alpha r^{2}$, so $e_{r}<\frac{1}{\alpha r^{2}}=\frac{1}{\alpha}\cdot\frac{1}{r^{2}}$. By Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §comparison, Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §distributive, clause reciprocal-squares and $0<\frac{1}{\alpha}$,

$$\sum_{r=1}^{M}e_{r}\le\frac{1}{\alpha}\sum_{r=1}^{M}\frac{1}{r^{2}}\le\frac{1}{\alpha}\Big(2-\frac{1}{M}\Big)\le\frac{2}{\alpha}.$$

Therefore $\sum_{m=1}^{2M+1}c_{m}\le1+\frac{4}{\alpha}$.

Clause exponential. Put $A=|x|\ge0$ and $u_{k}=\frac{A^{k}}{k!}$ for $k\in\mathbb{N}_{0}$. By Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §sign, $u_{k}\ge0$ and $\big|\frac{x^{k}}{k!}\big|=\frac{|x^{k}|}{k!}=u_{k}$, as $k!>0$. By The Factorial: Recursion, Positivity and Bounds by Powers §recursion, Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product, $(k+1)!=(k+1)\,k!$ and $A^{k+1}=A^{k}A$, so $u_{k+1}=u_{k}\cdot\frac{A}{k+1}$ for $k\in\mathbb{N}_{0}$.

Choose, depending on $x$ only, $N\in\mathbb{N}$ with $2A<N$, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §archimedean. Let $Q$ be the class of those $p\in\mathbb{N}_{0}$ with $u_{N+p}\le u_{N}\big(\frac{1}{2}\big)^{p}$. Then $0\in Q$, as $N+0=N$ and $\big(\frac{1}{2}\big)^{0}=1$ (Commutative Rings, Fields and Ordered Fields: Standard Notation §rings). Let $p\in Q$ and $q=N+p+1$. Then $N\le N+(p+1)=q$, so $2A<N\le q$ in $\mathbb{R}$ and $\frac{A}{q}<\frac{1}{2}$; as $u_{N+p}\ge0$, $u_{N+p+1}=u_{N+p}\cdot\frac{A}{q}\le\frac{1}{2}u_{N+p}\le u_{N}\big(\frac{1}{2}\big)^{p+1}$, using Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §exponents and Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product. So $p+1\in Q$, and $Q=\mathbb{N}_{0}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §induction.

By Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series §geometric, applied with $r=\frac{1}{2}$ (as $|\frac{1}{2}|=\frac{1}{2}<1$), the series $\sum_{k=1}^{\infty}\big(\frac{1}{2}\big)^{k}$ converges with sum $\frac{1/2}{1-1/2}=1$; its terms are positive by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §sign, so its partial sums are at most $1$ by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series §bounded. Hence for $n\in\mathbb{N}$, by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §comparison and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §distributive,

$$\sum_{k=1}^{n}u_{N+k}\le u_{N}\sum_{k=1}^{n}\Big(\frac{1}{2}\Big)^{k}\le u_{N}.$$

So the series $\sum_{k=1}^{\infty}u_{N+k}$ has nonnegative terms and partial sums bounded above by $u_{N}$, and it converges by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series §bounded and Bounded Sequences of Real Numbers §bounded. By Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping §shift with $p=N$, $\sum_{k=1}^{\infty}u_{k}$ converges; let $S$ be its sum. As $\big|\frac{x^{k}}{k!}\big|=u_{k}$, the series $\sum_{k=1}^{\infty}\frac{x^{k}}{k!}$ converges absolutely.

Now let $\varphi:\mathbb{N}\to\mathbb{N}$ be either $k\mapsto2k$ or $k\mapsto2k+1$. It is injective, by cancellation in $\mathbb{N}$, and for $n\in\mathbb{N}$ and $k\in[n]$ one has $1\le k\le2k\le\varphi(k)\le2k+1\le2n+1$, as $1\le k\le n$. So $\varphi$ maps $[n]$ bijectively onto $E_{n}=\{\varphi(k):k\in[n]\}\subseteq[2n+1]$, and by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §reindexing, Sums over Finite Sets in a Commutative Ring and in an Ordered Field: Distributivity, Products of Sums, Vanishing Terms, Sums over Pairs, Expanding Products of Sums, Counting, Comparison, Monotonicity and the Triangle Inequality §monotone and Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series §bounded,

$$\sum_{k=1}^{n}u_{\varphi(k)}=\sum_{j\in E_{n}}u_{j}\le\sum_{j=1}^{2n+1}u_{j}\le S.$$

Thus $\sum_{k=1}^{\infty}u_{\varphi(k)}$ has nonnegative terms and partial sums bounded above by $S$, so it converges by Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series §bounded and Bounded Sequences of Real Numbers §bounded. Since $\big|\frac{x^{\varphi(k)}}{\varphi(k)!}\big|=u_{\varphi(k)}$, the series $\sum_{k=1}^{\infty}\frac{x^{2k}}{(2k)!}$ and $\sum_{k=1}^{\infty}\frac{x^{2k+1}}{(2k+1)!}$ converge absolutely.