(* Title: HOL/ex/HarmonicSeries.thy
Author: Benjamin Porter, 2006
*)
section \<open>Divergence of the Harmonic Series\<close>
theory HarmonicSeries
imports Complex_Main
begin
subsection \<open>Abstract\<close>
text \<open>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.
\<close>
subsection \<open>Formal Proof\<close>
lemma two_pow_sub:
"0 < m \ (2::nat)^m - 2^(m - 1) = 2^(m - 1)"
by (induct m) auto
text \<open>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 $\frac{1}{3} +
\frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.\<close>
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))*(\<Sum>n\<in>(?S m). 1)"
by simp
also have
"\ = ((1/(real tm)) * real (card (?S m)))"
by (simp add: real_of_card)
also have
"\ = ((1/(real tm)) * real (tm - (2^(m - 1))))"
by (simp add: tmdef)
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"
by (simp add: tmdef)
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 \<open>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: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +
\frac{1}{6} + \frac{1}{7} + \frac{1}{8} = 1 + (\frac{1}{2}) +
(\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7}
+ \frac{1}{8})$.\<close>
lemma harmonic_aux2 [rule_format]:
"0 (\n\{1..(2::nat)^M}. 1/real n) =
(1 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n))"
(is "0 ?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 \<comment> \<open>show that LHS = c and RHS = c, and thus LHS = RHS\<close>
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)))"
by (subst sum.head)
(auto simp: atLeastSucAtMost_greaterThanAtMost)
also have
"\ = ((\n\{2..2::nat}. 1/real n) + 1/(real (1::nat)))"
by (simp add: eval_nat_numeral)
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) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1 / real n))"
proof -
have
"{1..(2::nat)^(Suc M)} =
{1..(2::nat)^M}\<union>{(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 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n) +
(\<Sum>n\<in>{(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 \<open>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.\<close>
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 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(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 = "(\x. 1/2)"
let ?g = "(\x. (\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 \<open>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.\<close>
theorem DivergenceOfHarmonicSeries:
shows "\summable (\n. 1/real (Suc n))"
(is "\summable ?f")
proof \<comment> \<open>by contradiction\<close>
let ?s = "suminf ?f" \<comment> \<open>let ?s equal the sum of the harmonic series\<close>
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 "(\i
using sf by (simp add: sum_less_suminf)
then have "(\i\{Suc 0..
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
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