src/HOL/ex/HarmonicSeries.thy
 author nipkow Fri Mar 06 17:38:47 2009 +0100 (2009-03-06) changeset 30313 b2441b0c8d38 parent 28952 15a4b2cf8c34 child 33509 29e4cf2c4ea3 permissions -rw-r--r--
     1 (*  Title:      HOL/ex/HarmonicSeries.thy

     2     Author:     Benjamin Porter, 2006

     3 *)

     4

     5 header {* Divergence of the Harmonic Series *}

     6

     7 theory HarmonicSeries

     8 imports Complex_Main

     9 begin

    10

    11 section {* Abstract *}

    12

    13 text {* 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 *}

    37

    38 section {* Formal Proof *}

    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 {* 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 $\frac{1}{3} +   50 \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$. *}

    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 with 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

    76         "inverse (real x) \<ge> inverse (real ((2::nat)^m))" by simp

    77       moreover

    78       from xgt0 have "real x \<noteq> 0" by simp

    79       then have

    80         "inverse (real x) = 1 / (real x)"

    81         by (rule nonzero_inverse_eq_divide)

    82       moreover from mgt0 have "real tm \<noteq> 0" by (simp add: tmdef)

    83       then have

    84         "inverse (real tm) = 1 / (real tm)"

    85         by (rule nonzero_inverse_eq_divide)

    86       ultimately show

    87         "1/(real x) \<ge> 1/(real tm)" by (auto simp add: tmdef)

    88     qed

    89     then have

    90       "(\<Sum>n\<in>(?S m). 1 / real n) \<ge> (\<Sum>n\<in>(?S m). 1/(real tm))"

    91       by (rule setsum_mono)

    92     moreover have

    93       "(\<Sum>n\<in>(?S m). 1/(real tm)) = 1/2"

    94     proof -

    95       have

    96         "(\<Sum>n\<in>(?S m). 1/(real tm)) =

    97          (1/(real tm))*(\<Sum>n\<in>(?S m). 1)"

    98         by simp

    99       also have

   100         "\<dots> = ((1/(real tm)) * real (card (?S m)))"

   101         by (simp add: real_of_card real_of_nat_def)

   102       also have

   103         "\<dots> = ((1/(real tm)) * real (tm - (2^(m - 1))))"

   104         by (simp add: tmdef)

   105       also from mgt0 have

   106         "\<dots> = ((1/(real tm)) * real ((2::nat)^(m - 1)))"

   107         by (auto simp: tmdef dest: two_pow_sub)

   108       also have

   109         "\<dots> = (real (2::nat))^(m - 1) / (real (2::nat))^m"

   110         by (simp add: tmdef realpow_real_of_nat [symmetric])

   111       also from mgt0 have

   112         "\<dots> = (real (2::nat))^(m - 1) / (real (2::nat))^((m - 1) + 1)"

   113         by auto

   114       also have "\<dots> = 1/2" by simp

   115       finally show ?thesis .

   116     qed

   117     ultimately have

   118       "(\<Sum>n\<in>(?S m). 1 / real n) \<ge> 1/2"

   119       by - (erule subst)

   120   }

   121   thus "0 < m \<longrightarrow> 1 / 2 \<le> (\<Sum>n\<in>(?S m). 1 / real n)" by simp

   122 qed

   123

   124 text {* We then show that the sum of a finite number of terms from the

   125 harmonic series can be regrouped in increasing powers of 2. For

   126 example: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +   127 \frac{1}{6} + \frac{1}{7} + \frac{1}{8} = 1 + (\frac{1}{2}) +   128 (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7}   129 + \frac{1}{8})$. *}

   130

   131 lemma harmonic_aux2 [rule_format]:

   132   "0<M \<Longrightarrow> (\<Sum>n\<in>{1..(2::nat)^M}. 1/real n) =

   133    (1 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n))"

   134   (is "0<M \<Longrightarrow> ?LHS M = ?RHS M")

   135 proof (induct M)

   136   case 0 show ?case by simp

   137 next

   138   case (Suc M)

   139   have ant: "0 < Suc M" by fact

   140   {

   141     have suc: "?LHS (Suc M) = ?RHS (Suc M)"

   142     proof cases -- "show that LHS = c and RHS = c, and thus LHS = RHS"

   143       assume mz: "M=0"

   144       {

   145         then have

   146           "?LHS (Suc M) = ?LHS 1" by simp

   147         also have

   148           "\<dots> = (\<Sum>n\<in>{(1::nat)..2}. 1/real n)" by simp

   149         also have

   150           "\<dots> = ((\<Sum>n\<in>{Suc 1..2}. 1/real n) + 1/(real (1::nat)))"

   151           by (subst setsum_head)

   152              (auto simp: atLeastSucAtMost_greaterThanAtMost)

   153         also have

   154           "\<dots> = ((\<Sum>n\<in>{2..2::nat}. 1/real n) + 1/(real (1::nat)))"

   155           by (simp add: nat_number)

   156         also have

   157           "\<dots> =  1/(real (2::nat)) + 1/(real (1::nat))" by simp

   158         finally have

   159           "?LHS (Suc M) = 1/2 + 1" by simp

   160       }

   161       moreover

   162       {

   163         from mz have

   164           "?RHS (Suc M) = ?RHS 1" by simp

   165         also have

   166           "\<dots> = (\<Sum>n\<in>{((2::nat)^0)+1..2^1}. 1/real n) + 1"

   167           by simp

   168         also have

   169           "\<dots> = (\<Sum>n\<in>{2::nat..2}. 1/real n) + 1"

   170         proof -

   171           have "(2::nat)^0 = 1" by simp

   172           then have "(2::nat)^0+1 = 2" by simp

   173           moreover have "(2::nat)^1 = 2" by simp

   174           ultimately have "{((2::nat)^0)+1..2^1} = {2::nat..2}" by auto

   175           thus ?thesis by simp

   176         qed

   177         also have

   178           "\<dots> = 1/2 + 1"

   179           by simp

   180         finally have

   181           "?RHS (Suc M) = 1/2 + 1" by simp

   182       }

   183       ultimately show "?LHS (Suc M) = ?RHS (Suc M)" by simp

   184     next

   185       assume mnz: "M\<noteq>0"

   186       then have mgtz: "M>0" by simp

   187       with Suc have suc:

   188         "(?LHS M) = (?RHS M)" by blast

   189       have

   190         "(?LHS (Suc M)) =

   191          ((?LHS M) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1 / real n))"

   192       proof -

   193         have

   194           "{1..(2::nat)^(Suc M)} =

   195            {1..(2::nat)^M}\<union>{(2::nat)^M+1..(2::nat)^(Suc M)}"

   196           by auto

   197         moreover have

   198           "{1..(2::nat)^M}\<inter>{(2::nat)^M+1..(2::nat)^(Suc M)} = {}"

   199           by auto

   200         moreover have

   201           "finite {1..(2::nat)^M}" and "finite {(2::nat)^M+1..(2::nat)^(Suc M)}"

   202           by auto

   203         ultimately show ?thesis

   204           by (auto intro: setsum_Un_disjoint)

   205       qed

   206       moreover

   207       {

   208         have

   209           "(?RHS (Suc M)) =

   210            (1 + (\<Sum>m\<in>{1..M}.  \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n) +

   211            (\<Sum>n\<in>{(2::nat)^(Suc M - 1)+1..2^(Suc M)}. 1/real n))" by simp

   212         also have

   213           "\<dots> = (?RHS M) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1/real n)"

   214           by simp

   215         also from suc have

   216           "\<dots> = (?LHS M) +  (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1/real n)"

   217           by simp

   218         finally have

   219           "(?RHS (Suc M)) = \<dots>" by simp

   220       }

   221       ultimately show "?LHS (Suc M) = ?RHS (Suc M)" by simp

   222     qed

   223   }

   224   thus ?case by simp

   225 qed

   226

   227 text {* Using @{thm [source] harmonic_aux} and @{thm [source] harmonic_aux2} we now show

   228 that each group sum is greater than or equal to $\frac{1}{2}$ and thus

   229 the finite sum is bounded below by a value proportional to the number

   230 of elements we choose. *}

   231

   232 lemma harmonic_aux3 [rule_format]:

   233   shows "\<forall>(M::nat). (\<Sum>n\<in>{1..(2::nat)^M}. 1 / real n) \<ge> 1 + (real M)/2"

   234   (is "\<forall>M. ?P M \<ge> _")

   235 proof (rule allI, cases)

   236   fix M::nat

   237   assume "M=0"

   238   then show "?P M \<ge> 1 + (real M)/2" by simp

   239 next

   240   fix M::nat

   241   assume "M\<noteq>0"

   242   then have "M > 0" by simp

   243   then have

   244     "(?P M) =

   245      (1 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n))"

   246     by (rule harmonic_aux2)

   247   also have

   248     "\<dots> \<ge> (1 + (\<Sum>m\<in>{1..M}. 1/2))"

   249   proof -

   250     let ?f = "(\<lambda>x. 1/2)"

   251     let ?g = "(\<lambda>x. (\<Sum>n\<in>{(2::nat)^(x - 1)+1..2^x}. 1/real n))"

   252     from harmonic_aux have "\<And>x. x\<in>{1..M} \<Longrightarrow> ?f x \<le> ?g x" by simp

   253     then have "(\<Sum>m\<in>{1..M}. ?g m) \<ge> (\<Sum>m\<in>{1..M}. ?f m)" by (rule setsum_mono)

   254     thus ?thesis by simp

   255   qed

   256   finally have "(?P M) \<ge> (1 + (\<Sum>m\<in>{1..M}. 1/2))" .

   257   moreover

   258   {

   259     have

   260       "(\<Sum>m\<in>{1..M}. (1::real)/2) = 1/2 * (\<Sum>m\<in>{1..M}. 1)"

   261       by auto

   262     also have

   263       "\<dots> = 1/2*(real (card {1..M}))"

   264       by (simp only: real_of_card[symmetric])

   265     also have

   266       "\<dots> = 1/2*(real M)" by simp

   267     also have

   268       "\<dots> = (real M)/2" by simp

   269     finally have "(\<Sum>m\<in>{1..M}. (1::real)/2) = (real M)/2" .

   270   }

   271   ultimately show "(?P M) \<ge> (1 + (real M)/2)" by simp

   272 qed

   273

   274 text {* The final theorem shows that as we take more and more elements

   275 (see @{thm [source] harmonic_aux3}) we get an ever increasing sum. By assuming

   276 the sum converges, the lemma @{thm [source] series_pos_less} ( @{thm

   277 series_pos_less} ) states that each sum is bounded above by the

   278 series' limit. This contradicts our first statement and thus we prove

   279 that the harmonic series is divergent. *}

   280

   281 theorem DivergenceOfHarmonicSeries:

   282   shows "\<not>summable (\<lambda>n. 1/real (Suc n))"

   283   (is "\<not>summable ?f")

   284 proof -- "by contradiction"

   285   let ?s = "suminf ?f" -- "let ?s equal the sum of the harmonic series"

   286   assume sf: "summable ?f"

   287   then obtain n::nat where ndef: "n = nat \<lceil>2 * ?s\<rceil>" by simp

   288   then have ngt: "1 + real n/2 > ?s"

   289   proof -

   290     have "\<forall>n. 0 \<le> ?f n" by simp

   291     with sf have "?s \<ge> 0"

   292       by - (rule suminf_0_le, simp_all)

   293     then have cgt0: "\<lceil>2*?s\<rceil> \<ge> 0" by simp

   294

   295     from ndef have "n = nat \<lceil>(2*?s)\<rceil>" .

   296     then have "real n = real (nat \<lceil>2*?s\<rceil>)" by simp

   297     with cgt0 have "real n = real \<lceil>2*?s\<rceil>"

   298       by (auto dest: real_nat_eq_real)

   299     then have "real n \<ge> 2*(?s)" by simp

   300     then have "real n/2 \<ge> (?s)" by simp

   301     then show "1 + real n/2 > (?s)" by simp

   302   qed

   303

   304   obtain j where jdef: "j = (2::nat)^n" by simp

   305   have "\<forall>m\<ge>j. 0 < ?f m" by simp

   306   with sf have "(\<Sum>i\<in>{0..<j}. ?f i) < ?s" by (rule series_pos_less)

   307   then have "(\<Sum>i\<in>{1..<Suc j}. 1/(real i)) < ?s"

   308     apply -

   309     apply (subst(asm) setsum_shift_bounds_Suc_ivl [symmetric])

   310     by simp

   311   with jdef have

   312     "(\<Sum>i\<in>{1..< Suc ((2::nat)^n)}. 1 / (real i)) < ?s" by simp

   313   then have

   314     "(\<Sum>i\<in>{1..(2::nat)^n}. 1 / (real i)) < ?s"

   315     by (simp only: atLeastLessThanSuc_atLeastAtMost)

   316   moreover from harmonic_aux3 have

   317     "(\<Sum>i\<in>{1..(2::nat)^n}. 1 / (real i)) \<ge> 1 + real n/2" by simp

   318   moreover from ngt have "1 + real n/2 > ?s" by simp

   319   ultimately show False by simp

   320 qed

   321

   322 end