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+++ b/src/HOL/ex/HarmonicSeries.thy Wed Dec 03 15:58:44 2008 +0100
@@ -0,0 +1,322 @@
+(* Title: HOL/ex/HarmonicSeries.thy
+ Author: Benjamin Porter, 2006
+*)
+
+header {* Divergence of the Harmonic Series *}
+
+theory HarmonicSeries
+imports Complex_Main
+begin
+
+section {* 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.
+*}
+
+section {* Formal Proof *}
+
+lemma two_pow_sub:
+ "0 < m \<Longrightarrow> (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:
+ "\<forall>m>0. (\<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n) \<ge> 1/2"
+ (is "\<forall>m>0. (\<Sum>n\<in>(?S m). 1/real n) \<ge> 1/2")
+proof
+ fix m::nat
+ obtain tm where tmdef: "tm = (2::nat)^m" by simp
+ {
+ assume mgt0: "0 < m"
+ have "\<And>x. x\<in>(?S m) \<Longrightarrow> 1/(real x) \<ge> 1/(real tm)"
+ proof -
+ fix x::nat
+ assume xs: "x\<in>(?S m)"
+ have xgt0: "x>0"
+ proof -
+ from xs have
+ "x \<ge> 2^(m - 1) + 1" by auto
+ moreover with mgt0 have
+ "2^(m - 1) + 1 \<ge> (1::nat)" by auto
+ ultimately have
+ "x \<ge> 1" by (rule xtrans)
+ thus ?thesis by simp
+ qed
+ moreover from xs have "x \<le> 2^m" by auto
+ ultimately have
+ "inverse (real x) \<ge> inverse (real ((2::nat)^m))" by simp
+ moreover
+ from xgt0 have "real x \<noteq> 0" by simp
+ then have
+ "inverse (real x) = 1 / (real x)"
+ by (rule nonzero_inverse_eq_divide)
+ moreover from mgt0 have "real tm \<noteq> 0" by (simp add: tmdef)
+ then have
+ "inverse (real tm) = 1 / (real tm)"
+ by (rule nonzero_inverse_eq_divide)
+ ultimately show
+ "1/(real x) \<ge> 1/(real tm)" by (auto simp add: tmdef)
+ qed
+ then have
+ "(\<Sum>n\<in>(?S m). 1 / real n) \<ge> (\<Sum>n\<in>(?S m). 1/(real tm))"
+ by (rule setsum_mono)
+ moreover have
+ "(\<Sum>n\<in>(?S m). 1/(real tm)) = 1/2"
+ proof -
+ have
+ "(\<Sum>n\<in>(?S m). 1/(real tm)) =
+ (1/(real tm))*(\<Sum>n\<in>(?S m). 1)"
+ by simp
+ also have
+ "\<dots> = ((1/(real tm)) * real (card (?S m)))"
+ by (simp add: real_of_card real_of_nat_def)
+ also have
+ "\<dots> = ((1/(real tm)) * real (tm - (2^(m - 1))))"
+ by (simp add: tmdef)
+ also from mgt0 have
+ "\<dots> = ((1/(real tm)) * real ((2::nat)^(m - 1)))"
+ by (auto simp: tmdef dest: two_pow_sub)
+ also have
+ "\<dots> = (real (2::nat))^(m - 1) / (real (2::nat))^m"
+ by (simp add: tmdef realpow_real_of_nat [symmetric])
+ also from mgt0 have
+ "\<dots> = (real (2::nat))^(m - 1) / (real (2::nat))^((m - 1) + 1)"
+ by auto
+ also have "\<dots> = 1/2" by simp
+ finally show ?thesis .
+ qed
+ ultimately have
+ "(\<Sum>n\<in>(?S m). 1 / real n) \<ge> 1/2"
+ by - (erule subst)
+ }
+ thus "0 < m \<longrightarrow> 1 / 2 \<le> (\<Sum>n\<in>(?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 \<Longrightarrow> (\<Sum>n\<in>{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<M \<Longrightarrow> ?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
+ "\<dots> = (\<Sum>n\<in>{(1::nat)..2}. 1/real n)" by simp
+ also have
+ "\<dots> = ((\<Sum>n\<in>{Suc 1..2}. 1/real n) + 1/(real (1::nat)))"
+ by (subst setsum_head)
+ (auto simp: atLeastSucAtMost_greaterThanAtMost)
+ also have
+ "\<dots> = ((\<Sum>n\<in>{2..2::nat}. 1/real n) + 1/(real (1::nat)))"
+ by (simp add: nat_number)
+ also have
+ "\<dots> = 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
+ "\<dots> = (\<Sum>n\<in>{((2::nat)^0)+1..2^1}. 1/real n) + 1"
+ by simp
+ also have
+ "\<dots> = (\<Sum>n\<in>{2::nat..2}. 1/real n) + 1"
+ proof -
+ have "(2::nat)^0 = 1" by simp
+ then have "(2::nat)^0+1 = 2" by simp
+ moreover have "(2::nat)^1 = 2" by simp
+ ultimately have "{((2::nat)^0)+1..2^1} = {2::nat..2}" by auto
+ thus ?thesis by simp
+ qed
+ also have
+ "\<dots> = 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\<noteq>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}\<inter>{(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: setsum_Un_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
+ "\<dots> = (?RHS M) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1/real n)"
+ by simp
+ also from suc have
+ "\<dots> = (?LHS M) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1/real n)"
+ by simp
+ finally have
+ "(?RHS (Suc M)) = \<dots>" 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 "\<forall>(M::nat). (\<Sum>n\<in>{1..(2::nat)^M}. 1 / real n) \<ge> 1 + (real M)/2"
+ (is "\<forall>M. ?P M \<ge> _")
+proof (rule allI, cases)
+ fix M::nat
+ assume "M=0"
+ then show "?P M \<ge> 1 + (real M)/2" by simp
+next
+ fix M::nat
+ assume "M\<noteq>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
+ "\<dots> \<ge> (1 + (\<Sum>m\<in>{1..M}. 1/2))"
+ proof -
+ let ?f = "(\<lambda>x. 1/2)"
+ let ?g = "(\<lambda>x. (\<Sum>n\<in>{(2::nat)^(x - 1)+1..2^x}. 1/real n))"
+ from harmonic_aux have "\<And>x. x\<in>{1..M} \<Longrightarrow> ?f x \<le> ?g x" by simp
+ then have "(\<Sum>m\<in>{1..M}. ?g m) \<ge> (\<Sum>m\<in>{1..M}. ?f m)" by (rule setsum_mono)
+ thus ?thesis by simp
+ qed
+ finally have "(?P M) \<ge> (1 + (\<Sum>m\<in>{1..M}. 1/2))" .
+ moreover
+ {
+ have
+ "(\<Sum>m\<in>{1..M}. (1::real)/2) = 1/2 * (\<Sum>m\<in>{1..M}. 1)"
+ by auto
+ also have
+ "\<dots> = 1/2*(real (card {1..M}))"
+ by (simp only: real_of_card[symmetric])
+ also have
+ "\<dots> = 1/2*(real M)" by simp
+ also have
+ "\<dots> = (real M)/2" by simp
+ finally have "(\<Sum>m\<in>{1..M}. (1::real)/2) = (real M)/2" .
+ }
+ ultimately show "(?P M) \<ge> (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] series_pos_less} ( @{thm
+series_pos_less} ) 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 "\<not>summable (\<lambda>n. 1/real (Suc n))"
+ (is "\<not>summable ?f")
+proof -- "by contradiction"
+ 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 \<lceil>2 * ?s\<rceil>" by simp
+ then have ngt: "1 + real n/2 > ?s"
+ proof -
+ have "\<forall>n. 0 \<le> ?f n" by simp
+ with sf have "?s \<ge> 0"
+ by - (rule suminf_0_le, simp_all)
+ then have cgt0: "\<lceil>2*?s\<rceil> \<ge> 0" by simp
+
+ from ndef have "n = nat \<lceil>(2*?s)\<rceil>" .
+ then have "real n = real (nat \<lceil>2*?s\<rceil>)" by simp
+ with cgt0 have "real n = real \<lceil>2*?s\<rceil>"
+ by (auto dest: real_nat_eq_real)
+ then have "real n \<ge> 2*(?s)" by simp
+ then have "real n/2 \<ge> (?s)" by simp
+ then show "1 + real n/2 > (?s)" by simp
+ qed
+
+ obtain j where jdef: "j = (2::nat)^n" by simp
+ have "\<forall>m\<ge>j. 0 < ?f m" by simp
+ with sf have "(\<Sum>i\<in>{0..<j}. ?f i) < ?s" by (rule series_pos_less)
+ then have "(\<Sum>i\<in>{1..<Suc j}. 1/(real i)) < ?s"
+ apply -
+ apply (subst(asm) setsum_shift_bounds_Suc_ivl [symmetric])
+ by simp
+ with jdef have
+ "(\<Sum>i\<in>{1..< Suc ((2::nat)^n)}. 1 / (real i)) < ?s" by simp
+ then have
+ "(\<Sum>i\<in>{1..(2::nat)^n}. 1 / (real i)) < ?s"
+ by (simp only: atLeastLessThanSuc_atLeastAtMost)
+ moreover from harmonic_aux3 have
+ "(\<Sum>i\<in>{1..(2::nat)^n}. 1 / (real i)) \<ge> 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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