# HG changeset patch
# User kleing
# Date 1140351692 -3600
# Node ID 6e6b5b1fdc06a9bb4a5af4958f4bcc63232688e0
# Parent  3aabd46340e03f2cef73af398f64da83289789cd
* added Library/ASeries (sum of arithmetic series with instantiation to nat and int)
* added Complex/ex/ASeries_Complex (instantiation of the above for reals)
* added Complex/ex/HarmonicSeries (should really be in something like Complex/Library)

(these are contributions by Benjamin Porter, numbers 68 and 34 of
http://www.cs.ru.nl/~freek/100/)

diff -r 3aabd46340e0 -r 6e6b5b1fdc06 src/HOL/Complex/ex/ASeries_Complex.thy
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/src/HOL/Complex/ex/ASeries_Complex.thy	Sun Feb 19 13:21:32 2006 +0100
@@ -0,0 +1,24 @@
+(*  Title:      HOL/Library/ASeries.thy
+    ID:         $Id$
+    Author:     Benjamin Porter, 2006
+*)
+
+
+header {* Arithmetic Series for Reals *}
+
+theory ASeries_Complex
+imports Complex_Main ASeries
+begin
+
+lemma arith_series_real:
+  "(2::real) * (\<Sum>i\<in>{..<n}. a + of_nat i * d) =
+  of_nat n * (a + (a + of_nat(n - 1)*d))"
+proof -
+  have
+    "((1::real) + 1) * (\<Sum>i\<in>{..<n}. a + of_nat(i)*d) =
+    of_nat(n) * (a + (a + of_nat(n - 1)*d))"
+    by (rule arith_series_general)
+  thus ?thesis by simp
+qed
+
+end
diff -r 3aabd46340e0 -r 6e6b5b1fdc06 src/HOL/Complex/ex/HarmonicSeries.thy
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/src/HOL/Complex/ex/HarmonicSeries.thy	Sun Feb 19 13:21:32 2006 +0100
@@ -0,0 +1,323 @@
+(*  Title:      HOL/Library/HarmonicSeries.thy
+    ID:         $Id$
+    Author:     Benjamin Porter, 2006
+*)
+
+header {* Divergence of the Harmonic Series *}
+
+theory HarmonicSeries
+imports Complex_Main
+begin
+
+section {* 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.
+*}
+
+section {* Formal Proof *}
+
+lemma two_pow_sub:
+  "0 < m \<Longrightarrow> (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:
+  "\<forall>m>0. (\<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n) \<ge> 1/2"
+  (is "\<forall>m>0. (\<Sum>n\<in>(?S m). 1/real n) \<ge> 1/2")
+proof
+  fix m::nat
+  obtain tm where tmdef: "tm = (2::nat)^m" by simp
+  {
+    assume mgt0: "0 < m"
+    have "\<And>x. x\<in>(?S m) \<Longrightarrow> 1/(real x) \<ge> 1/(real tm)"
+    proof -
+      fix x::nat
+      assume xs: "x\<in>(?S m)"
+      have xgt0: "x>0"
+      proof -
+        from xs have
+          "x \<ge> 2^(m - 1) + 1" by auto
+        moreover with mgt0 have
+          "2^(m - 1) + 1 \<ge> (1::nat)" by auto
+        ultimately have
+          "x \<ge> 1" by (rule xtrans)
+        thus ?thesis by simp
+      qed
+      moreover from xs have "x \<le> 2^m" by auto
+      ultimately have
+        "inverse (real x) \<ge> inverse (real ((2::nat)^m))" by simp
+      moreover
+      from xgt0 have "real x \<noteq> 0" by simp
+      then have
+        "inverse (real x) = 1 / (real x)"
+        by (rule nonzero_inverse_eq_divide)
+      moreover from mgt0 have "real tm \<noteq> 0" by (simp add: tmdef)
+      then have
+        "inverse (real tm) = 1 / (real tm)"
+        by (rule nonzero_inverse_eq_divide)
+      ultimately show
+        "1/(real x) \<ge> 1/(real tm)" by (auto simp add: tmdef)
+    qed
+    then have
+      "(\<Sum>n\<in>(?S m). 1 / real n) \<ge> (\<Sum>n\<in>(?S m). 1/(real tm))"
+      by (rule setsum_mono)
+    moreover have
+      "(\<Sum>n\<in>(?S m). 1/(real tm)) = 1/2"
+    proof -
+      have
+        "(\<Sum>n\<in>(?S m). 1/(real tm)) =
+         (1/(real tm))*(\<Sum>n\<in>(?S m). 1)"
+        by simp
+      also have
+        "\<dots> = ((1/(real tm)) * real (card (?S m)))"
+        by (simp add: real_of_card real_of_nat_def)
+      also have
+        "\<dots> = ((1/(real tm)) * real (tm - (2^(m - 1))))"
+        by (simp add: tmdef)
+      also from mgt0 have
+        "\<dots> = ((1/(real tm)) * real ((2::nat)^(m - 1)))"
+        by (auto simp: tmdef dest: two_pow_sub)
+      also have
+        "\<dots> = (real (2::nat))^(m - 1) / (real (2::nat))^m"
+        by (simp add: tmdef realpow_real_of_nat [symmetric])
+      also from mgt0 have
+        "\<dots> = (real (2::nat))^(m - 1) / (real (2::nat))^((m - 1) + 1)"
+        by auto
+      also have "\<dots> = 1/2" by simp
+      finally show ?thesis .
+    qed
+    ultimately have
+      "(\<Sum>n\<in>(?S m). 1 / real n) \<ge> 1/2"
+      by - (erule subst)
+  }
+  thus "0 < m \<longrightarrow> 1 / 2 \<le> (\<Sum>n\<in>(?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 \<Longrightarrow> (\<Sum>n\<in>{1..(2::nat)^M}. 1/real n) =
+   (1 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n))"
+  (is "0<M \<Longrightarrow> ?LHS M = ?RHS M")
+proof (induct M)
+  case 0 show ?case by simp
+next
+  case (Suc M)
+  have ant: "0 < Suc M" .
+  {
+    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
+          "\<dots> = (\<Sum>n\<in>{(1::nat)..2}. 1/real n)" by simp
+        also have
+          "\<dots> = ((\<Sum>n\<in>{Suc 1..2}. 1/real n) + 1/(real (1::nat)))"
+          by (subst setsum_head)
+             (auto simp: atLeastSucAtMost_greaterThanAtMost)
+        also have
+          "\<dots> = ((\<Sum>n\<in>{2..2::nat}. 1/real n) + 1/(real (1::nat)))"
+          by (simp add: nat_number)
+        also have
+          "\<dots> =  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
+          "\<dots> = (\<Sum>n\<in>{((2::nat)^0)+1..2^1}. 1/real n) + 1"
+          by simp
+        also have
+          "\<dots> = (\<Sum>n\<in>{2::nat..2}. 1/real n) + 1"
+        proof -
+          have "(2::nat)^0 = 1" by simp
+          then have "(2::nat)^0+1 = 2" by simp
+          moreover have "(2::nat)^1 = 2" by simp
+          ultimately have "{((2::nat)^0)+1..2^1} = {2::nat..2}" by auto
+          thus ?thesis by simp
+        qed
+        also have
+          "\<dots> = 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\<noteq>0"
+      then have mgtz: "M>0" by simp
+      with Suc have suc:
+        "(?LHS M) = (?RHS M)" by blast
+      have
+        "(?LHS (Suc M)) =
+         ((?LHS M) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1 / real n))"
+      proof -
+        have
+          "{1..(2::nat)^(Suc M)} =
+           {1..(2::nat)^M}\<union>{(2::nat)^M+1..(2::nat)^(Suc M)}"
+          by auto
+        moreover have
+          "{1..(2::nat)^M}\<inter>{(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 + (\<Sum>m\<in>{1..M}.  \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n) +
+           (\<Sum>n\<in>{(2::nat)^(Suc M - 1)+1..2^(Suc M)}. 1/real n))" by simp
+        also have
+          "\<dots> = (?RHS M) + (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1/real n)"
+          by simp
+        also from suc have
+          "\<dots> = (?LHS M) +  (\<Sum>n\<in>{(2::nat)^M+1..2^(Suc M)}. 1/real n)"
+          by simp
+        finally have
+          "(?RHS (Suc M)) = \<dots>" 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 "\<forall>(M::nat). (\<Sum>n\<in>{1..(2::nat)^M}. 1 / real n) \<ge> 1 + (real M)/2"
+  (is "\<forall>M. ?P M \<ge> _")
+proof (rule allI, cases)
+  fix M::nat
+  assume "M=0"
+  then show "?P M \<ge> 1 + (real M)/2" by simp
+next
+  fix M::nat
+  assume "M\<noteq>0"
+  then have "M > 0" by simp
+  then have
+    "(?P M) =
+     (1 + (\<Sum>m\<in>{1..M}. \<Sum>n\<in>{(2::nat)^(m - 1)+1..2^m}. 1/real n))"
+    by (rule harmonic_aux2)
+  also have
+    "\<dots> \<ge> (1 + (\<Sum>m\<in>{1..M}. 1/2))"
+  proof -
+    let ?f = "(\<lambda>x. 1/2)"
+    let ?g = "(\<lambda>x. (\<Sum>n\<in>{(2::nat)^(x - 1)+1..2^x}. 1/real n))"
+    from harmonic_aux have "\<And>x. x\<in>{1..M} \<Longrightarrow> ?f x \<le> ?g x" by simp
+    then have "(\<Sum>m\<in>{1..M}. ?g m) \<ge> (\<Sum>m\<in>{1..M}. ?f m)" by (rule setsum_mono)
+    thus ?thesis by simp
+  qed
+  finally have "(?P M) \<ge> (1 + (\<Sum>m\<in>{1..M}. 1/2))" .
+  moreover
+  {
+    have
+      "(\<Sum>m\<in>{1..M}. (1::real)/2) = 1/2 * (\<Sum>m\<in>{1..M}. 1)"
+      by auto
+    also have
+      "\<dots> = 1/2*(real (card {1..M}))"
+      by (simp only: real_of_card[symmetric])
+    also have
+      "\<dots> = 1/2*(real M)" by simp
+    also have
+      "\<dots> = (real M)/2" by simp
+    finally have "(\<Sum>m\<in>{1..M}. (1::real)/2) = (real M)/2" .
+  }
+  ultimately show "(?P M) \<ge> (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] series_pos_less} ( @{thm
+series_pos_less} ) 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 "\<not>summable (\<lambda>n. 1/real (Suc n))"
+  (is "\<not>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 \<lceil>2 * ?s\<rceil>" by simp
+  then have ngt: "1 + real n/2 > ?s"
+  proof -
+    have "\<forall>n. 0 \<le> ?f n" by simp
+    with sf have "?s \<ge> 0"
+      by - (rule suminf_0_le, simp_all)
+    then have cgt0: "\<lceil>2*?s\<rceil> \<ge> 0" by simp
+
+    from ndef have "n = nat \<lceil>(2*?s)\<rceil>" .
+    then have "real n = real (nat \<lceil>2*?s\<rceil>)" by simp
+    with cgt0 have "real n = real \<lceil>2*?s\<rceil>"
+      by (auto dest: real_nat_eq_real)
+    then have "real n \<ge> 2*(?s)" by simp
+    then have "real n/2 \<ge> (?s)" by simp
+    then show "1 + real n/2 > (?s)" by simp
+  qed
+
+  obtain j where jdef: "j = (2::nat)^n" by simp
+  have "\<forall>m\<ge>j. 0 < ?f m" by simp
+  with sf have "(\<Sum>i\<in>{0..<j}. ?f i) < ?s" by (rule series_pos_less)
+  then have "(\<Sum>i\<in>{1..<Suc j}. 1/(real i)) < ?s"
+    apply -
+    apply (subst(asm) setsum_shift_bounds_Suc_ivl [symmetric])
+    by simp
+  with jdef have
+    "(\<Sum>i\<in>{1..< Suc ((2::nat)^n)}. 1 / (real i)) < ?s" by simp
+  then have
+    "(\<Sum>i\<in>{1..(2::nat)^n}. 1 / (real i)) < ?s"
+    by (simp only: atLeastLessThanSuc_atLeastAtMost)
+  moreover from harmonic_aux3 have
+    "(\<Sum>i\<in>{1..(2::nat)^n}. 1 / (real i)) \<ge> 1 + real n/2" by simp
+  moreover from ngt have "1 + real n/2 > ?s" by simp
+  ultimately show False by simp
+qed
+
+end
\ No newline at end of file
diff -r 3aabd46340e0 -r 6e6b5b1fdc06 src/HOL/Complex/ex/ROOT.ML
--- a/src/HOL/Complex/ex/ROOT.ML	Sun Feb 19 02:11:27 2006 +0100
+++ b/src/HOL/Complex/ex/ROOT.ML	Sun Feb 19 13:21:32 2006 +0100
@@ -14,3 +14,6 @@
 
 no_document use_thy "BigO";
 use_thy "BigO_Complex";
+
+use_thy "ASeries_Complex";
+use_thy "HarmonicSeries";
diff -r 3aabd46340e0 -r 6e6b5b1fdc06 src/HOL/Hyperreal/Series.thy
--- a/src/HOL/Hyperreal/Series.thy	Sun Feb 19 02:11:27 2006 +0100
+++ b/src/HOL/Hyperreal/Series.thy	Sun Feb 19 13:21:32 2006 +0100
@@ -280,7 +280,6 @@
 apply simp
 done
 
-
 lemma sumr_pos_lt_pair:
      "[|summable f; 
         \<forall>d. 0 < (f(n + (Suc(Suc 0) * d))) + f(n + ((Suc(Suc 0) * d) + 1))|]  
@@ -409,6 +408,21 @@
 apply (simp add: abs_le_interval_iff)
 done
 
+(* specialisation for the common 0 case *)
+lemma suminf_0_le:
+  fixes f::"nat\<Rightarrow>real"
+  assumes gt0: "\<forall>n. 0 \<le> f n" and sm: "summable f"
+  shows "0 \<le> suminf f"
+proof -
+  let ?g = "(\<lambda>n. (0::real))"
+  from gt0 have "\<forall>n. ?g n \<le> f n" by simp
+  moreover have "summable ?g" by (rule summable_zero)
+  moreover from sm have "summable f" .
+  ultimately have "suminf ?g \<le> suminf f" by (rule summable_le)
+  then show "0 \<le> suminf f" by (simp add: suminf_zero)
+qed 
+
+
 text{*Absolute convergence imples normal convergence*}
 lemma summable_rabs_cancel: "summable (%n. abs (f n)) ==> summable f"
 apply (auto simp add: summable_Cauchy)
diff -r 3aabd46340e0 -r 6e6b5b1fdc06 src/HOL/Library/ASeries.thy
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/src/HOL/Library/ASeries.thy	Sun Feb 19 13:21:32 2006 +0100
@@ -0,0 +1,114 @@
+(*  Title:      HOL/Library/ASeries.thy
+    ID:         $Id$
+    Author:     Benjamin Porter, 2006
+*)
+
+
+header {* Arithmetic Series *}
+
+theory ASeries
+imports Main
+begin
+
+section {* Abstract *}
+
+text {* The following document presents a proof of the Arithmetic
+Series Sum formalised in Isabelle/Isar.
+
+{\em Theorem:} The series $\sum_{i=1}^{n} a_i$ where $a_{i+1} = a_i +
+d$ for some constant $d$ has the sum $\frac{n}{2} (a_1 + a_n)$
+(i.e. $n$ multiplied by the arithmetic mean of the first and last
+element).
+
+{\em Informal Proof:} (from
+"http://mathworld.wolfram.com/ArithmeticSeries.html")
+  The proof is a simple forward proof. Let $S$ equal the sum above and
+  $a$ the first element, then we have
+\begin{align}
+\nonumber  S &= a + (a+d) + (a+2d) + ... a_n \\
+\nonumber    &= n*a + d (0 + 1 + 2 + ... n-1) \\
+\nonumber    &= n*a + d (\frac{1}{2} * (n-1) * n)    \mbox{..using a simple sum identity} \\
+\nonumber    &= \frac{n}{2} (2a + d(n-1)) \\
+\nonumber    & \mbox{..but $(a+a_n = a + (a + d(n-1)) = 2a + d(n-1))$ so} \\
+\nonumber  S &= \frac{n}{2} (a + a_n)
+\end{align}
+*}
+
+section {* Formal Proof *}
+
+text {* We present a proof for the abstract case of a commutative ring,
+we then instantiate for three common types nats, ints and reals. The
+function @{text "of_nat"} maps the natural numbers into any
+commutative ring.
+*}
+
+lemmas comm_simp [simp] = left_distrib right_distrib add_assoc mult_ac
+
+text {* Next we prove the following simple summation law $\sum_{i=1}^n
+i = \frac {n * (n+1)}{2}$. *}
+
+lemma sum_ident:
+  "((1::'a::comm_semiring_1_cancel) + 1)*(\<Sum>i\<in>{1..n}. of_nat i) =
+   of_nat n*((of_nat n)+1)"
+proof (induct n)
+  case 0
+  show ?case by simp
+next
+  case (Suc n)
+  then show ?case by simp
+qed
+
+text {* The abstract theorem follows. Note that $2$ is displayed as
+$1+1$ to keep the structure as abstract as possible. *}
+
+theorem arith_series_general:
+  "((1::'a::comm_semiring_1_cancel) + 1) * (\<Sum>i\<in>{..<n}. a + of_nat i * d) =
+  of_nat n * (a + (a + of_nat(n - 1)*d))"
+proof cases
+  assume ngt1: "n > 1"
+  let ?I = "\<lambda>i. of_nat i" and ?n = "of_nat n"
+  have
+    "(\<Sum>i\<in>{..<n}. a+?I i*d) =
+     ((\<Sum>i\<in>{..<n}. a) + (\<Sum>i\<in>{..<n}. ?I i*d))"
+    by (rule setsum_addf)
+  also from ngt1 have "\<dots> = ?n*a + (\<Sum>i\<in>{..<n}. ?I i*d)" by simp
+  also from ngt1 have "\<dots> = (?n*a + d*(\<Sum>i\<in>{1..<n}. ?I i))"
+    by (simp add: setsum_mult setsum_head_upt)
+  also have "(1+1)*\<dots> = (1+1)*?n*a + d*(1+1)*(\<Sum>i\<in>{1..<n}. ?I i)"
+    by simp
+  also from ngt1 have "{1..<n} = {1..n - 1}"
+    by (cases n) (auto simp: atLeastLessThanSuc_atLeastAtMost)    
+  also from ngt1 
+  have "(1+1)*?n*a + d*(1+1)*(\<Sum>i\<in>{1..n - 1}. ?I i) = ((1+1)*?n*a + d*?I (n - 1)*?I n)"
+    by (simp only: mult_ac sum_ident [of "n - 1"]) (simp add: of_nat_Suc [symmetric])
+  finally show ?thesis by simp
+next
+  assume "\<not>(n > 1)"
+  hence "n = 1 \<or> n = 0" by auto
+  thus ?thesis by auto
+qed
+
+subsection {* Instantiation *}
+
+lemma arith_series_nat:
+  "(2::nat) * (\<Sum>i\<in>{..<n}. a+i*d) = n * (a + (a+(n - 1)*d))"
+proof -
+  have
+    "((1::nat) + 1) * (\<Sum>i\<in>{..<n::nat}. a + of_nat(i)*d) =
+    of_nat(n) * (a + (a + of_nat(n - 1)*d))"
+    by (rule arith_series_general)
+  thus ?thesis by (auto simp add: of_nat_id)
+qed
+
+lemma arith_series_int:
+  "(2::int) * (\<Sum>i\<in>{..<n}. a + of_nat i * d) =
+  of_nat n * (a + (a + of_nat(n - 1)*d))"
+proof -
+  have
+    "((1::int) + 1) * (\<Sum>i\<in>{..<n}. a + of_nat i * d) =
+    of_nat(n) * (a + (a + of_nat(n - 1)*d))"
+    by (rule arith_series_general)
+  thus ?thesis by simp
+qed
+
+end
diff -r 3aabd46340e0 -r 6e6b5b1fdc06 src/HOL/Library/Library.thy
--- a/src/HOL/Library/Library.thy	Sun Feb 19 02:11:27 2006 +0100
+++ b/src/HOL/Library/Library.thy	Sun Feb 19 13:21:32 2006 +0100
@@ -21,6 +21,7 @@
   Char_ord
   Commutative_Ring
   Coinductive_List
+  ASeries
 begin
 end
 (*>*)
diff -r 3aabd46340e0 -r 6e6b5b1fdc06 src/HOL/SetInterval.thy
--- a/src/HOL/SetInterval.thy	Sun Feb 19 02:11:27 2006 +0100
+++ b/src/HOL/SetInterval.thy	Sun Feb 19 13:21:32 2006 +0100
@@ -715,24 +715,39 @@
   "setsum f {Suc m..<Suc n} = setsum (%i. f(Suc i)){m..<n}"
 by (simp add:setsum_shift_bounds_nat_ivl[where k=1,simplified])
 
-lemma setsum_rmv_head:
+lemma setsum_head:
+  fixes n :: nat
+  assumes mn: "m <= n" 
+  shows "(\<Sum>x\<in>{m..n}. P x) = P m + (\<Sum>x\<in>{m<..n}. P x)" (is "?lhs = ?rhs")
+proof -
+  from mn
+  have "{m..n} = {m} \<union> {m<..n}"
+    by (auto intro: ivl_disj_un_singleton)
+  hence "?lhs = (\<Sum>x\<in>{m} \<union> {m<..n}. P x)"
+    by (simp add: atLeast0LessThan)
+  also have "\<dots> = ?rhs" by simp
+  finally show ?thesis .
+qed
+
+lemma setsum_head_upt:
   fixes m::nat
   assumes m: "0 < m"
-  shows "P 0 + (\<Sum>x\<in>{1..<m}. P x) = (\<Sum>x<m. P x)"
-  (is "?lhs = ?rhs")
+  shows "(\<Sum>x<m. P x) = P 0 + (\<Sum>x\<in>{1..<m}. P x)"
 proof -
-  let ?a = "\<Sum>x\<in>({0} \<union> {0<..<m}). P x"
-  from m
-  have "{0..<m} = {0} \<union> {0<..<m}"
-    by (simp only: ivl_disj_un_singleton)
-  hence "?rhs = ?a"
+  have "(\<Sum>x<m. P x) = (\<Sum>x\<in>{0..<m}. P x)" 
     by (simp add: atLeast0LessThan)
-  moreover
-  have "?a = ?lhs"
-    by (simp add: setsum_Un ivl_disj_int 
-                  atLeastSucLessThan_greaterThanLessThan)
-  ultimately 
-  show ?thesis by simp
+  also 
+  from m 
+  have "\<dots> = (\<Sum>x\<in>{0..m - 1}. P x)"
+    by (cases m) (auto simp add: atLeastLessThanSuc_atLeastAtMost)
+  also
+  have "\<dots> = P 0 + (\<Sum>x\<in>{0<..m - 1}. P x)"
+    by (simp add: setsum_head)
+  also 
+  from m 
+  have "{0<..m - 1} = {1..<m}" 
+    by (cases m) (auto simp add: atLeastLessThanSuc_atLeastAtMost)
+  finally show ?thesis .
 qed
 
 subsection {* The formula for geometric sums *}
diff -r 3aabd46340e0 -r 6e6b5b1fdc06 src/HOL/ex/ThreeDivides.thy
--- a/src/HOL/ex/ThreeDivides.thy	Sun Feb 19 02:11:27 2006 +0100
+++ b/src/HOL/ex/ThreeDivides.thy	Sun Feb 19 13:21:32 2006 +0100
@@ -207,7 +207,7 @@
          (simp add: atLeast0LessThan)
     also have
       "\<dots> = (\<Sum>x<Suc nd. m div 10^x mod 10 * 10^x)"
-      by (simp add: setsum_rmv_head [symmetric] cdef)
+      by (simp add: setsum_rmv_upt cdef)
     also note `Suc nd = nlen m`
     finally
     show "m = (\<Sum>x<nlen m. m div 10^x mod 10 * 10^x)" .