# Theory HarmonicSeries

theory HarmonicSeries
imports Complex_Main
(*  Title:      HOL/ex/HarmonicSeries.thy
Author:     Benjamin Porter, 2006
*)

section ‹Divergence of the Harmonic Series›

theory HarmonicSeries
imports Complex_Main
begin

subsection ‹Abstract›

text ‹The following document presents a proof of the Divergence of
Harmonic Series theorem formalised in the Isabelle/Isar theorem
proving system.

{\em Theorem:} The series $\sum_{n=1}^{\infty} \frac{1}{n}$ does not
converge to any number.

{\em Informal Proof:}
The informal proof is based on the following auxillary lemmas:
\begin{itemize}
\item{aux: $\sum_{n=2^m-1}^{2^m} \frac{1}{n} \geq \frac{1}{2}$}
\item{aux2: $\sum_{n=1}^{2^M} \frac{1}{n} = 1 + \sum_{m=1}^{M} \sum_{n=2^m-1}^{2^m} \frac{1}{n}$}
\end{itemize}

From {\em aux} and {\em aux2} we can deduce that $\sum_{n=1}^{2^M} \frac{1}{n} \geq 1 + \frac{M}{2}$ for all $M$.
Now for contradiction, assume that $\sum_{n=1}^{\infty} \frac{1}{n} = s$ for some $s$. Because $\forall n. \frac{1}{n} > 0$ all the
partial sums in the series must be less than $s$. However with our
deduction above we can choose $N > 2*s - 2$ and thus
$\sum_{n=1}^{2^N} \frac{1}{n} > s$. This leads to a contradiction
and hence $\sum_{n=1}^{\infty} \frac{1}{n}$ is not summable.
QED.
›

subsection ‹Formal Proof›

lemma two_pow_sub:
"0 < m ⟹ (2::nat)^m - 2^(m - 1) = 2^(m - 1)"
by (induct m) auto

text ‹We first prove the following auxillary lemma. This lemma
simply states that the finite sums: $\frac{1}{2}$, $\frac{1}{3} + \frac{1}{4}$, $\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}$
etc. are all greater than or equal to $\frac{1}{2}$. We do this by
observing that each term in the sum is greater than or equal to the
last term, e.g. $\frac{1}{3} > \frac{1}{4}$ and thus $\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.›

lemma harmonic_aux:
"∀m>0. (∑n∈{(2::nat)^(m - 1)+1..2^m}. 1/real n) ≥ 1/2"
(is "∀m>0. (∑n∈(?S m). 1/real n) ≥ 1/2")
proof
fix m::nat
obtain tm where tmdef: "tm = (2::nat)^m" by simp
{
assume mgt0: "0 < m"
have "⋀x. x∈(?S m) ⟹ 1/(real x) ≥ 1/(real tm)"
proof -
fix x::nat
assume xs: "x∈(?S m)"
have xgt0: "x>0"
proof -
from xs have
"x ≥ 2^(m - 1) + 1" by auto
moreover from mgt0 have
"2^(m - 1) + 1 ≥ (1::nat)" by auto
ultimately have
"x ≥ 1" by (rule xtrans)
thus ?thesis by simp
qed
moreover from xs have "x ≤ 2^m" by auto
ultimately have "inverse (real x) ≥ inverse (real ((2::nat)^m))" by simp
moreover
from xgt0 have "real x ≠ 0" by simp
then have
"inverse (real x) = 1 / (real x)"
by (rule nonzero_inverse_eq_divide)
moreover from mgt0 have "real tm ≠ 0" by (simp add: tmdef)
then have
"inverse (real tm) = 1 / (real tm)"
by (rule nonzero_inverse_eq_divide)
ultimately show
"1/(real x) ≥ 1/(real tm)" by (auto simp add: tmdef)
qed
then have
"(∑n∈(?S m). 1 / real n) ≥ (∑n∈(?S m). 1/(real tm))"
by (rule sum_mono)
moreover have
"(∑n∈(?S m). 1/(real tm)) = 1/2"
proof -
have
"(∑n∈(?S m). 1/(real tm)) =
(1/(real tm))*(∑n∈(?S m). 1)"
by simp
also have
"… = ((1/(real tm)) * real (card (?S m)))"
also have
"… = ((1/(real tm)) * real (tm - (2^(m - 1))))"
also from mgt0 have
"… = ((1/(real tm)) * real ((2::nat)^(m - 1)))"
by (auto simp: tmdef dest: two_pow_sub)
also have
"… = (real (2::nat))^(m - 1) / (real (2::nat))^m"
also from mgt0 have
"… = (real (2::nat))^(m - 1) / (real (2::nat))^((m - 1) + 1)"
by auto
also have "… = 1/2" by simp
finally show ?thesis .
qed
ultimately have
"(∑n∈(?S m). 1 / real n) ≥ 1/2"
by - (erule subst)
}
thus "0 < m ⟶ 1 / 2 ≤ (∑n∈(?S m). 1 / real n)" by simp
qed

text ‹We then show that the sum of a finite number of terms from the
harmonic series can be regrouped in increasing powers of 2. For
example: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} = 1 + (\frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8})$.›

lemma harmonic_aux2 [rule_format]:
"0<M ⟹ (∑n∈{1..(2::nat)^M}. 1/real n) =
(1 + (∑m∈{1..M}. ∑n∈{(2::nat)^(m - 1)+1..2^m}. 1/real n))"
(is "0<M ⟹ ?LHS M = ?RHS M")
proof (induct M)
case 0 show ?case by simp
next
case (Suc M)
have ant: "0 < Suc M" by fact
{
have suc: "?LHS (Suc M) = ?RHS (Suc M)"
proof cases ― ‹show that LHS = c and RHS = c, and thus LHS = RHS›
assume mz: "M=0"
{
then have
"?LHS (Suc M) = ?LHS 1" by simp
also have
"… = (∑n∈{(1::nat)..2}. 1/real n)" by simp
also have
"… = ((∑n∈{Suc 1..2}. 1/real n) + 1/(real (1::nat)))"
(auto simp: atLeastSucAtMost_greaterThanAtMost)
also have
"… = ((∑n∈{2..2::nat}. 1/real n) + 1/(real (1::nat)))"
also have
"… =  1/(real (2::nat)) + 1/(real (1::nat))" by simp
finally have
"?LHS (Suc M) = 1/2 + 1" by simp
}
moreover
{
from mz have
"?RHS (Suc M) = ?RHS 1" by simp
also have
"… = (∑n∈{((2::nat)^0)+1..2^1}. 1/real n) + 1"
by simp
also have
"… = (∑n∈{2::nat..2}. 1/real n) + 1"
by (auto simp: atLeastAtMost_singleton')
also have
"… = 1/2 + 1"
by simp
finally have
"?RHS (Suc M) = 1/2 + 1" by simp
}
ultimately show "?LHS (Suc M) = ?RHS (Suc M)" by simp
next
assume mnz: "M≠0"
then have mgtz: "M>0" by simp
with Suc have suc:
"(?LHS M) = (?RHS M)" by blast
have
"(?LHS (Suc M)) =
((?LHS M) + (∑n∈{(2::nat)^M+1..2^(Suc M)}. 1 / real n))"
proof -
have
"{1..(2::nat)^(Suc M)} =
{1..(2::nat)^M}∪{(2::nat)^M+1..(2::nat)^(Suc M)}"
by auto
moreover have
"{1..(2::nat)^M}∩{(2::nat)^M+1..(2::nat)^(Suc M)} = {}"
by auto
moreover have
"finite {1..(2::nat)^M}" and "finite {(2::nat)^M+1..(2::nat)^(Suc M)}"
by auto
ultimately show ?thesis
by (auto intro: sum.union_disjoint)
qed
moreover
{
have
"(?RHS (Suc M)) =
(1 + (∑m∈{1..M}.  ∑n∈{(2::nat)^(m - 1)+1..2^m}. 1/real n) +
(∑n∈{(2::nat)^(Suc M - 1)+1..2^(Suc M)}. 1/real n))" by simp
also have
"… = (?RHS M) + (∑n∈{(2::nat)^M+1..2^(Suc M)}. 1/real n)"
by simp
also from suc have
"… = (?LHS M) +  (∑n∈{(2::nat)^M+1..2^(Suc M)}. 1/real n)"
by simp
finally have
"(?RHS (Suc M)) = …" by simp
}
ultimately show "?LHS (Suc M) = ?RHS (Suc M)" by simp
qed
}
thus ?case by simp
qed

text ‹Using @{thm [source] harmonic_aux} and @{thm [source] harmonic_aux2} we now show
that each group sum is greater than or equal to $\frac{1}{2}$ and thus
the finite sum is bounded below by a value proportional to the number
of elements we choose.›

lemma harmonic_aux3 [rule_format]:
shows "∀(M::nat). (∑n∈{1..(2::nat)^M}. 1 / real n) ≥ 1 + (real M)/2"
(is "∀M. ?P M ≥ _")
proof (rule allI, cases)
fix M::nat
assume "M=0"
then show "?P M ≥ 1 + (real M)/2" by simp
next
fix M::nat
assume "M≠0"
then have "M > 0" by simp
then have
"(?P M) =
(1 + (∑m∈{1..M}. ∑n∈{(2::nat)^(m - 1)+1..2^m}. 1/real n))"
by (rule harmonic_aux2)
also have
"… ≥ (1 + (∑m∈{1..M}. 1/2))"
proof -
let ?f = "(λx. 1/2)"
let ?g = "(λx. (∑n∈{(2::nat)^(x - 1)+1..2^x}. 1/real n))"
from harmonic_aux have "⋀x. x∈{1..M} ⟹ ?f x ≤ ?g x" by simp
then have "(∑m∈{1..M}. ?g m) ≥ (∑m∈{1..M}. ?f m)" by (rule sum_mono)
thus ?thesis by simp
qed
finally have "(?P M) ≥ (1 + (∑m∈{1..M}. 1/2))" .
moreover
{
have
"(∑m∈{1..M}. (1::real)/2) = 1/2 * (∑m∈{1..M}. 1)"
by auto
also have
"… = 1/2*(real (card {1..M}))"
by (simp only: real_of_card[symmetric])
also have
"… = 1/2*(real M)" by simp
also have
"… = (real M)/2" by simp
finally have "(∑m∈{1..M}. (1::real)/2) = (real M)/2" .
}
ultimately show "(?P M) ≥ (1 + (real M)/2)" by simp
qed

text ‹The final theorem shows that as we take more and more elements
(see @{thm [source] harmonic_aux3}) we get an ever increasing sum. By assuming
the sum converges, the lemma @{thm [source] sum_less_suminf} ( @{thm
sum_less_suminf} ) states that each sum is bounded above by the
series' limit. This contradicts our first statement and thus we prove
that the harmonic series is divergent.›

theorem DivergenceOfHarmonicSeries:
shows "¬summable (λn. 1/real (Suc n))"
(is "¬summable ?f")
let ?s = "suminf ?f" ― ‹let ?s equal the sum of the harmonic series›
assume sf: "summable ?f"
then obtain n::nat where ndef: "n = nat ⌈2 * ?s⌉" by simp
then have ngt: "1 + real n/2 > ?s" by linarith
define j where "j = (2::nat)^n"
have "∀m≥j. 0 < ?f m" by simp
with sf have "(∑i<j. ?f i) < ?s" by (rule sum_less_suminf)
then have "(∑i∈{Suc 0..<Suc j}. 1/(real i)) < ?s"
unfolding sum_shift_bounds_Suc_ivl by (simp add: atLeast0LessThan)
with j_def have
"(∑i∈{1..< Suc ((2::nat)^n)}. 1 / (real i)) < ?s" by simp
then have
"(∑i∈{1..(2::nat)^n}. 1 / (real i)) < ?s"
by (simp only: atLeastLessThanSuc_atLeastAtMost)
moreover from harmonic_aux3 have
"(∑i∈{1..(2::nat)^n}. 1 / (real i)) ≥ 1 + real n/2" by simp
moreover from ngt have "1 + real n/2 > ?s" by simp
ultimately show False by simp
qed

end