(* Title: HOL/ex/HarmonicSeries.thy Author: Benjamin Porter, 2006 *) section ‹Divergence of the Harmonic Series› theory HarmonicSeries imports Complex_Main begin subsection ‹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. › subsection ‹Formal Proof› lemma two_pow_sub: "0 < m ⟹ (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: "∀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 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) 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 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 ⟹ (∑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 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 "… = (∑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.union_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 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 "∀(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 simp next 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 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] setsum_less_suminf} ( @{thm setsum_less_suminf} ) 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 "¬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" by linarith def j ≡ "(2::nat)^n" have "∀m≥j. 0 < ?f m" by simp with sf have "(∑i<j. ?f i) < ?s" by (rule setsum_less_suminf) then have "(∑i∈{Suc 0..<Suc j}. 1/(real i)) < ?s" unfolding setsum_shift_bounds_Suc_ivl by (simp add: atLeast0LessThan) with j_def 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 simp qed end