
diff -r 1860f016886d -r 15a4b2cf8c34 src/HOL/ex/HarmonicSeries.thy
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ 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
\ No newline at end of file




