src/HOL/ex/HarmonicSeries.thy
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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