src/HOL/ex/HarmonicSeries.thy
author wenzelm
Wed Jun 22 10:09:20 2016 +0200 (2016-06-22)
changeset 63343 fb5d8a50c641
parent 63040 eb4ddd18d635
child 64267 b9a1486e79be
permissions -rw-r--r--
bundle lifting_syntax;
     1 (*  Title:      HOL/ex/HarmonicSeries.thy
     2     Author:     Benjamin Porter, 2006
     3 *)
     4 
     5 section \<open>Divergence of the Harmonic Series\<close>
     6 
     7 theory HarmonicSeries
     8 imports Complex_Main
     9 begin
    10 
    11 subsection \<open>Abstract\<close>
    12 
    13 text \<open>The following document presents a proof of the Divergence of
    14 Harmonic Series theorem formalised in the Isabelle/Isar theorem
    15 proving system.
    16 
    17 {\em Theorem:} The series $\sum_{n=1}^{\infty} \frac{1}{n}$ does not
    18 converge to any number.
    19 
    20 {\em Informal Proof:}
    21   The informal proof is based on the following auxillary lemmas:
    22   \begin{itemize}
    23   \item{aux: $\sum_{n=2^m-1}^{2^m} \frac{1}{n} \geq \frac{1}{2}$}
    24   \item{aux2: $\sum_{n=1}^{2^M} \frac{1}{n} = 1 + \sum_{m=1}^{M} \sum_{n=2^m-1}^{2^m} \frac{1}{n}$}
    25   \end{itemize}
    26 
    27   From {\em aux} and {\em aux2} we can deduce that $\sum_{n=1}^{2^M}
    28   \frac{1}{n} \geq 1 + \frac{M}{2}$ for all $M$.
    29   Now for contradiction, assume that $\sum_{n=1}^{\infty} \frac{1}{n}
    30   = s$ for some $s$. Because $\forall n. \frac{1}{n} > 0$ all the
    31   partial sums in the series must be less than $s$. However with our
    32   deduction above we can choose $N > 2*s - 2$ and thus
    33   $\sum_{n=1}^{2^N} \frac{1}{n} > s$. This leads to a contradiction
    34   and hence $\sum_{n=1}^{\infty} \frac{1}{n}$ is not summable.
    35   QED.
    36 \<close>
    37 
    38 subsection \<open>Formal Proof\<close>
    39 
    40 lemma two_pow_sub:
    41   "0 < m \<Longrightarrow> (2::nat)^m - 2^(m - 1) = 2^(m - 1)"
    42   by (induct m) auto
    43 
    44 text \<open>We first prove the following auxillary lemma. This lemma
    45 simply states that the finite sums: $\frac{1}{2}$, $\frac{1}{3} +
    46 \frac{1}{4}$, $\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}$
    47 etc. are all greater than or equal to $\frac{1}{2}$. We do this by
    48 observing that each term in the sum is greater than or equal to the
    49 last term, e.g. $\frac{1}{3} > \frac{1}{4}$ and thus $\frac{1}{3} +
    50 \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.\<close>
    51 
    52 lemma harmonic_aux:
    53   "\<forall>m>0. (\<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n) \<ge> 1/2"
    54   (is "\<forall>m>0. (\<Sum>n\<in>(?S m). 1/real n) \<ge> 1/2")
    55 proof
    56   fix m::nat
    57   obtain tm where tmdef: "tm = (2::nat)^m" by simp
    58   {
    59     assume mgt0: "0 < m"
    60     have "\<And>x. x\<in>(?S m) \<Longrightarrow> 1/(real x) \<ge> 1/(real tm)"
    61     proof -
    62       fix x::nat
    63       assume xs: "x\<in>(?S m)"
    64       have xgt0: "x>0"
    65       proof -
    66         from xs have
    67           "x \<ge> 2^(m - 1) + 1" by auto
    68         moreover from mgt0 have
    69           "2^(m - 1) + 1 \<ge> (1::nat)" by auto
    70         ultimately have
    71           "x \<ge> 1" by (rule xtrans)
    72         thus ?thesis by simp
    73       qed
    74       moreover from xs have "x \<le> 2^m" by auto
    75       ultimately have "inverse (real x) \<ge> inverse (real ((2::nat)^m))" by simp
    76       moreover
    77       from xgt0 have "real x \<noteq> 0" by simp
    78       then have
    79         "inverse (real x) = 1 / (real x)"
    80         by (rule nonzero_inverse_eq_divide)
    81       moreover from mgt0 have "real tm \<noteq> 0" by (simp add: tmdef)
    82       then have
    83         "inverse (real tm) = 1 / (real tm)"
    84         by (rule nonzero_inverse_eq_divide)
    85       ultimately show
    86         "1/(real x) \<ge> 1/(real tm)" by (auto simp add: tmdef)
    87     qed
    88     then have
    89       "(\<Sum>n\<in>(?S m). 1 / real n) \<ge> (\<Sum>n\<in>(?S m). 1/(real tm))"
    90       by (rule setsum_mono)
    91     moreover have
    92       "(\<Sum>n\<in>(?S m). 1/(real tm)) = 1/2"
    93     proof -
    94       have
    95         "(\<Sum>n\<in>(?S m). 1/(real tm)) =
    96          (1/(real tm))*(\<Sum>n\<in>(?S m). 1)"
    97         by simp
    98       also have
    99         "\<dots> = ((1/(real tm)) * real (card (?S m)))"
   100         by (simp add: real_of_card)
   101       also have
   102         "\<dots> = ((1/(real tm)) * real (tm - (2^(m - 1))))"
   103         by (simp add: tmdef)
   104       also from mgt0 have
   105         "\<dots> = ((1/(real tm)) * real ((2::nat)^(m - 1)))"
   106         by (auto simp: tmdef dest: two_pow_sub)
   107       also have
   108         "\<dots> = (real (2::nat))^(m - 1) / (real (2::nat))^m"
   109         by (simp add: tmdef)
   110       also from mgt0 have
   111         "\<dots> = (real (2::nat))^(m - 1) / (real (2::nat))^((m - 1) + 1)"
   112         by auto
   113       also have "\<dots> = 1/2" by simp
   114       finally show ?thesis .
   115     qed
   116     ultimately have
   117       "(\<Sum>n\<in>(?S m). 1 / real n) \<ge> 1/2"
   118       by - (erule subst)
   119   }
   120   thus "0 < m \<longrightarrow> 1 / 2 \<le> (\<Sum>n\<in>(?S m). 1 / real n)" by simp
   121 qed
   122 
   123 text \<open>We then show that the sum of a finite number of terms from the
   124 harmonic series can be regrouped in increasing powers of 2. For
   125 example: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +
   126 \frac{1}{6} + \frac{1}{7} + \frac{1}{8} = 1 + (\frac{1}{2}) +
   127 (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7}
   128 + \frac{1}{8})$.\<close>
   129 
   130 lemma harmonic_aux2 [rule_format]:
   131   "0<M \<Longrightarrow> (\<Sum>n\<in>{1..(2::nat)^M}. 1/real n) =
   132    (1 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n))"
   133   (is "0<M \<Longrightarrow> ?LHS M = ?RHS M")
   134 proof (induct M)
   135   case 0 show ?case by simp
   136 next
   137   case (Suc M)
   138   have ant: "0 < Suc M" by fact
   139   {
   140     have suc: "?LHS (Suc M) = ?RHS (Suc M)"
   141     proof cases \<comment> "show that LHS = c and RHS = c, and thus LHS = RHS"
   142       assume mz: "M=0"
   143       {
   144         then have
   145           "?LHS (Suc M) = ?LHS 1" by simp
   146         also have
   147           "\<dots> = (\<Sum>n\<in>{(1::nat)..2}. 1/real n)" by simp
   148         also have
   149           "\<dots> = ((\<Sum>n\<in>{Suc 1..2}. 1/real n) + 1/(real (1::nat)))"
   150           by (subst setsum_head)
   151              (auto simp: atLeastSucAtMost_greaterThanAtMost)
   152         also have
   153           "\<dots> = ((\<Sum>n\<in>{2..2::nat}. 1/real n) + 1/(real (1::nat)))"
   154           by (simp add: eval_nat_numeral)
   155         also have
   156           "\<dots> =  1/(real (2::nat)) + 1/(real (1::nat))" by simp
   157         finally have
   158           "?LHS (Suc M) = 1/2 + 1" by simp
   159       }
   160       moreover
   161       {
   162         from mz have
   163           "?RHS (Suc M) = ?RHS 1" by simp
   164         also have
   165           "\<dots> = (\<Sum>n\<in>{((2::nat)^0)+1..2^1}. 1/real n) + 1"
   166           by simp
   167         also have
   168           "\<dots> = (\<Sum>n\<in>{2::nat..2}. 1/real n) + 1"
   169           by (auto simp: atLeastAtMost_singleton')
   170         also have
   171           "\<dots> = 1/2 + 1"
   172           by simp
   173         finally have
   174           "?RHS (Suc M) = 1/2 + 1" by simp
   175       }
   176       ultimately show "?LHS (Suc M) = ?RHS (Suc M)" by simp
   177     next
   178       assume mnz: "M\<noteq>0"
   179       then have mgtz: "M>0" by simp
   180       with Suc have suc:
   181         "(?LHS M) = (?RHS M)" by blast
   182       have
   183         "(?LHS (Suc M)) =
   184          ((?LHS M) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1 / real n))"
   185       proof -
   186         have
   187           "{1..(2::nat)^(Suc M)} =
   188            {1..(2::nat)^M}\<union>{(2::nat)^M+1..(2::nat)^(Suc M)}"
   189           by auto
   190         moreover have
   191           "{1..(2::nat)^M}\<inter>{(2::nat)^M+1..(2::nat)^(Suc M)} = {}"
   192           by auto
   193         moreover have
   194           "finite {1..(2::nat)^M}" and "finite {(2::nat)^M+1..(2::nat)^(Suc M)}"
   195           by auto
   196         ultimately show ?thesis
   197           by (auto intro: setsum.union_disjoint)
   198       qed
   199       moreover
   200       {
   201         have
   202           "(?RHS (Suc M)) =
   203            (1 + (\<Sum>m\<in>{1..M}.  \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n) +
   204            (\<Sum>n\<in>{(2::nat)^(Suc M - 1)+1..2^(Suc M)}. 1/real n))" by simp
   205         also have
   206           "\<dots> = (?RHS M) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1/real n)"
   207           by simp
   208         also from suc have
   209           "\<dots> = (?LHS M) +  (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1/real n)"
   210           by simp
   211         finally have
   212           "(?RHS (Suc M)) = \<dots>" by simp
   213       }
   214       ultimately show "?LHS (Suc M) = ?RHS (Suc M)" by simp
   215     qed
   216   }
   217   thus ?case by simp
   218 qed
   219 
   220 text \<open>Using @{thm [source] harmonic_aux} and @{thm [source] harmonic_aux2} we now show
   221 that each group sum is greater than or equal to $\frac{1}{2}$ and thus
   222 the finite sum is bounded below by a value proportional to the number
   223 of elements we choose.\<close>
   224 
   225 lemma harmonic_aux3 [rule_format]:
   226   shows "\<forall>(M::nat). (\<Sum>n\<in>{1..(2::nat)^M}. 1 / real n) \<ge> 1 + (real M)/2"
   227   (is "\<forall>M. ?P M \<ge> _")
   228 proof (rule allI, cases)
   229   fix M::nat
   230   assume "M=0"
   231   then show "?P M \<ge> 1 + (real M)/2" by simp
   232 next
   233   fix M::nat
   234   assume "M\<noteq>0"
   235   then have "M > 0" by simp
   236   then have
   237     "(?P M) =
   238      (1 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n))"
   239     by (rule harmonic_aux2)
   240   also have
   241     "\<dots> \<ge> (1 + (\<Sum>m\<in>{1..M}. 1/2))"
   242   proof -
   243     let ?f = "(\<lambda>x. 1/2)"
   244     let ?g = "(\<lambda>x. (\<Sum>n\<in>{(2::nat)^(x - 1)+1..2^x}. 1/real n))"
   245     from harmonic_aux have "\<And>x. x\<in>{1..M} \<Longrightarrow> ?f x \<le> ?g x" by simp
   246     then have "(\<Sum>m\<in>{1..M}. ?g m) \<ge> (\<Sum>m\<in>{1..M}. ?f m)" by (rule setsum_mono)
   247     thus ?thesis by simp
   248   qed
   249   finally have "(?P M) \<ge> (1 + (\<Sum>m\<in>{1..M}. 1/2))" .
   250   moreover
   251   {
   252     have
   253       "(\<Sum>m\<in>{1..M}. (1::real)/2) = 1/2 * (\<Sum>m\<in>{1..M}. 1)"
   254       by auto
   255     also have
   256       "\<dots> = 1/2*(real (card {1..M}))"
   257       by (simp only: real_of_card[symmetric])
   258     also have
   259       "\<dots> = 1/2*(real M)" by simp
   260     also have
   261       "\<dots> = (real M)/2" by simp
   262     finally have "(\<Sum>m\<in>{1..M}. (1::real)/2) = (real M)/2" .
   263   }
   264   ultimately show "(?P M) \<ge> (1 + (real M)/2)" by simp
   265 qed
   266 
   267 text \<open>The final theorem shows that as we take more and more elements
   268 (see @{thm [source] harmonic_aux3}) we get an ever increasing sum. By assuming
   269 the sum converges, the lemma @{thm [source] setsum_less_suminf} ( @{thm
   270 setsum_less_suminf} ) states that each sum is bounded above by the
   271 series' limit. This contradicts our first statement and thus we prove
   272 that the harmonic series is divergent.\<close>
   273 
   274 theorem DivergenceOfHarmonicSeries:
   275   shows "\<not>summable (\<lambda>n. 1/real (Suc n))"
   276   (is "\<not>summable ?f")
   277 proof \<comment> "by contradiction"
   278   let ?s = "suminf ?f" \<comment> "let ?s equal the sum of the harmonic series"
   279   assume sf: "summable ?f"
   280   then obtain n::nat where ndef: "n = nat \<lceil>2 * ?s\<rceil>" by simp
   281   then have ngt: "1 + real n/2 > ?s" by linarith
   282   define j where "j = (2::nat)^n"
   283   have "\<forall>m\<ge>j. 0 < ?f m" by simp
   284   with sf have "(\<Sum>i<j. ?f i) < ?s" by (rule setsum_less_suminf)
   285   then have "(\<Sum>i\<in>{Suc 0..<Suc j}. 1/(real i)) < ?s"
   286     unfolding setsum_shift_bounds_Suc_ivl by (simp add: atLeast0LessThan)
   287   with j_def have
   288     "(\<Sum>i\<in>{1..< Suc ((2::nat)^n)}. 1 / (real i)) < ?s" by simp
   289   then have
   290     "(\<Sum>i\<in>{1..(2::nat)^n}. 1 / (real i)) < ?s"
   291     by (simp only: atLeastLessThanSuc_atLeastAtMost)
   292   moreover from harmonic_aux3 have
   293     "(\<Sum>i\<in>{1..(2::nat)^n}. 1 / (real i)) \<ge> 1 + real n/2" by simp
   294   moreover from ngt have "1 + real n/2 > ?s" by simp
   295   ultimately show False by simp
   296 qed
   297 
   298 end