# Theory HarmonicSeries

theory HarmonicSeries
imports Complex_Main
(*  Title:      HOL/ex/HarmonicSeries.thy    Author:     Benjamin Porter, 2006*)header {* Divergence of the Harmonic Series *}theory HarmonicSeriesimports Complex_Mainbeginsubsection {* Abstract *}text {* The following document presents a proof of the Divergence ofHarmonic Series theorem formalised in the Isabelle/Isar theoremproving system.{\em Theorem:} The series $\sum_{n=1}^{\infty} \frac{1}{n}$ does notconverge 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) autotext {* We first prove the following auxillary lemma. This lemmasimply 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 byobserving that each term in the sum is greater than or equal to thelast 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 del: real_of_nat_power)      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 setsum_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)))"        by (simp add: real_of_card real_of_nat_def)      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 simpqedtext {* We then show that the sum of a finite number of terms from theharmonic series can be regrouped in increasing powers of 2. Forexample: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +\frac{1}{6} + \frac{1}{7} + \frac{1}{8} = 1 + (\frac{1}{2}) +(\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7}+ \frac{1}{8})$. *}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 simpnext  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)))"          by (subst setsum_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) + (∑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: setsum_Un_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 simpqedtext {* Using @{thm [source] harmonic_aux} and @{thm [source] harmonic_aux2} we now showthat each group sum is greater than or equal to $\frac{1}{2}$ and thusthe finite sum is bounded below by a value proportional to the numberof 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 simpnext  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 setsum_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 simpqedtext {* The final theorem shows that as we take more and more elements(see @{thm [source] harmonic_aux3}) we get an ever increasing sum. By assumingthe sum converges, the lemma @{thm [source] series_pos_less} ( @{thmseries_pos_less} ) states that each sum is bounded above by theseries' limit. This contradicts our first statement and thus we provethat the harmonic series is divergent. *}theorem DivergenceOfHarmonicSeries:  shows "¬summable (λn. 1/real (Suc n))"  (is "¬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 ⌈2 * ?s⌉" by simp  then have ngt: "1 + real n/2 > ?s"  proof -    have "∀n. 0 ≤ ?f n" by simp    with sf have "?s ≥ 0"      by - (rule suminf_0_le, simp_all)    then have cgt0: "⌈2*?s⌉ ≥ 0" by simp    from ndef have "n = nat ⌈(2*?s)⌉" .    then have "real n = real (nat ⌈2*?s⌉)" by simp    with cgt0 have "real n = real ⌈2*?s⌉"      by (auto dest: real_nat_eq_real)    then have "real n ≥ 2*(?s)" by simp    then have "real n/2 ≥ (?s)" by simp    then show "1 + real n/2 > (?s)" by simp  qed  obtain j where jdef: "j = (2::nat)^n" by simp  have "∀m≥j. 0 < ?f m" by simp  with sf have "(∑i∈{0..<j}. ?f i) < ?s" by (rule series_pos_less)  then have "(∑i∈{1..<Suc j}. 1/(real i)) < ?s"    apply -    apply (subst(asm) setsum_shift_bounds_Suc_ivl [symmetric])    by simp  with jdef 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 simpqedend