4. Sequences and sums
4.4. Recitation #3: Sums, products, binomial coefficients
= 0 for any k ∈ {0, 1, . . . ,n} satisfyingk > n/2) and thus can be discarded; nevertheless it is easier to have the sum end at k=n rather than figure out where exactly its nonzero addends stop.
Hints to the proof of Proposition 4.3.20. This is probably the hardest statement in this section, which is not saying much. There is a proof by strong induction on n (using Theorem 4.3.7); it is not very enlightening and somewhat laborious due to the fiddling involved in making all the sums have the same upper limit. (See [Vorobi02, §15] or [AndCri17, Problem 2.4] for a sketch of this proof, but notice that both of these sources are sloppy with the summation limits.) We will state a more general result (Proposition 4.9.18) later on, at which point we will give this proof in full detail; Proposition 4.3.20 is merely a particular case of this latter result.
Alternatively, there is a combinatorial proof, which proceeds as follows: WLOG assume thatn≥1 (else, just check it by hand). Let[n−1]denote the set{1, 2, . . . ,n−1}.
, as can easily be proved by induction on n (or even by a nicer, combinatorial argument: see [19fco, Proposition 1.4.10]). Hence, the total number of all lacunar subsets of[n−1] is ∑n
k=0
n−k k
. Comparing these two results yields the claim of Proposition 4.3.20. (See [19fco, §1.4.5, proof of Proposition 1.3.32] for this proof.)
4.4. Recitation #3: Sums, products, binomial coefficients
Next comes an exercise in finite sums that is rather similar to Exercise 2.2.1:
Exercise 4.4.1. Let (f0, f1, f2, . . .)be the Fibonacci sequence. Prove that each inte-gern ≥0 satisfies
f1+ f3+ f5+· · ·+f2n−1= f2n.
Discussion of Exercise 4.4.1. This is easy to prove by induction onn, but let us try to prove it by the telescope principle.
Let n ≥ 0 be an integer. For each integeri ≥ 1, we have fi+1 = fi+ fi−1 (by the recursive definition of the Fibonacci numbers) and therefore
fi = fi+1− fi−1. (126)
Thus, for each positive integers, we have
f2s−1 = f(2s−1)+1− f(2s−1)−1 (by (126), applied to i=2s−1)
= f2s− f2(s−1) (127)
(since(2s−1) +1 =2sand (2s−1)−1=2(s−1)). Now, f1+ f3+ f5+· · ·+ f2n−1 =
∑
n s=1f2s−1
| {z }
=f2s−f2(s−1) (by (127))
=
∑
n s=1
f2s− f2(s−1)
= f2n− f2(1−1)
(by (74), applied to u=1, v=n and as = f2s)
= f2n−0 since f2(1−1) = f0=0
= f2n. This solves Exercise 4.4.1.
Exercise 4.4.1 has other solutions. For example, we may solve it (for n ≥ 1) by noticing that
f1
|{z}=1
+ f3
|{z}
=f2+f1
=f1+f2
+ f5
|{z}
=f4+f3
=f3+f4
+· · ·+ f2n−1
| {z }
=f2n−2+f2n−3
=f2n−3+f2n−2
=1+ (f1+ f2) + (f3+ f4) +· · ·+ (f2n−3+ f2n−2)
| {z }
=f1+f2+···+f2n−2
=f(2n−2)+2−1
(by Exercise 2.2.1, applied to 2n−2 instead ofn)
=1+ f(2n−2)+2−1= f(2n−2)+2 = f2n. (The case n=0 is easily checked manually.)
The next exercise appears, e.g., in [AndTet18, Introduction]:
Exercise 4.4.2. ”Simplify” n
−1 k∏=0
1+a2k
for a 6= 1 and n ∈ N. (The answer should not contain any∏ signs.)
Discussion of Exercise 4.4.2. Let abe a number such thata 6=1. Letn ∈ N. For each k ∈N, we have
1− a2k+1
| {z }
=a2k·2 (since 2k+1=2k·2)
=1− a2k·2
|{z}
=a2k2
=1−a2k2
=1−a2k 1+a2k
(128) (by the classical formula 1−b2 = (1−b) (1+b), applied tob =a2k) and therefore
1+a2k = 1−a2k+1
1−a2k . (129)
Now,
n−1 k
∏
=0
1+a2k
| {z }
=1−a2k+1 1−a2k
(by (129))
=
n−1 k
∏
=01−a2k+1 1−a2k =
∏
n s=11−a2s 1−a2s−1
(here, we have substituted s−1 for kin the product)
= 1−a2n 1−a21−1
by (109), applied to u=1, v=nand as =a2s
= 1−a2n 1−a
since a21−1 = a20 = a1 =a . So we found our answer:
n−1 k
∏
=0
1+a2k
= 1−a2n
1−a . (130)
But wait – did you spot the subtle error?
The error is the following: We divided the equality (128) by 1−a2k to get (129), but 1−a2k can be zero. Namely, ifa =−1, then 1−a2k will be zero for everyk >0;
thus, in all these cases, we have divided by zero. Nevertheless, our final result (130) holds even in these cases – it is only the proof that went wrong.
Fortunately, now that we know the equality (130), it is utterly straightforward to prove it by induction on n (using (128) in the induction step).76 Thus, while the
76Here is this proof in detail:
Proof of (130): Forget that we fixedn. We shall prove (130) by induction onn.
argument by which we originally arrived at (130) was flawed, we nevertheless were able to put it to good use, since it helped us find the formula (130), which we were then able to prove by a different (perfectly sound) argument.
This was a typical example of a phenomenon that is often observed in mathe-matics (particularly in modern research): Results are often discovered by means of reasoning that is not quite rigorous (to the point that it often sounds like quackery or cannot be communicated at all); then, one is forced to come up with different arguments to find valid proofs for the results. In our case, we did not have to look far (in essence, the induction proof was just a restatement of the not-quite-correct telescope argument avoiding unnecessary divisions, which helped it avoid division by zero); but this sort of “parallel construction” of proofs can be quite a long and difficult process.
The next exercises are about binomial coefficients:
Exercise 4.4.3. Show that
Induction step: Let m∈N. Assume (as the induction hypothesis) that (130) holds forn =m.
We must prove that (130) holds forn=m+1.
Our induction hypothesis says that (130) holds for n = m; in other words, it says that
m−1∏
In other words, (130) holds for n = m+1. This completes the induction step. Thus, (130) is proved.
Discussion of Exercise 4.4.3. Intuitively, the idea is clear from a look at Pascal’s tri-angle: Pascal’s triangle is symmetric across the vertical axis (by Theorem 4.3.10).
Thus, if n ∈ N is odd, then the n-th row of Pascal’s triangle (i.e., the row that begins with 1 and n) splits neatly into two equal halves (one half being to the left of the vertical axis, and the other half being to the right of it), and therefore the sum of the left half of this row is exactly 1
2 of the sum of all entries in this row; but the latter sum is 2n according to Corollary 4.3.17. Thus, the former sum is 2n−1.
Here is a formal way to restate this argument: Let n ∈ N be odd. Thus, n =
Dividing both sides of this equality by 2, we find 22u =
∑
u k=0n k
=
(n−1)/2 k
∑
=0n k
(sinceu = (n−1)/2). Hence,
(n−1)/2 k
∑
=0n k
=22u =2n−1 (since 2u=n−1). This solves Exercise 4.4.3.
Exercise 4.4.4. Recall once again the Fibonacci sequence (f0, f1, f2, . . .), which is defined recursively by f0 =0, f1=1, and
fn = fn−1+ fn−2 for all n≥2. (133) Now, let us define fn fornegative integersn as well, by “applying (133) back-wards”: This means that we set fn−2 = fn − fn−1 for all integers n ≤ 1. This allows us to recursively compute f−1, f−2, f−3, . . . (in this order). For example,
f−1 = f1− f0 =1−0=1;
f−2 = f0− f−1 =0−1=−1;
f−3 = f−1− f−2=1−(−1) =2, etc.
(a)Prove that f−n = (−1)n−1 fn for eachn ∈Z.
(b)Prove that fn+m+1 = fnfm+ fn+1fm+1 for all n∈ Zand m∈ Z.
(c)Prove that 7fn = fn−4+ fn+4for all n ∈Z.
(d)Prove that if a,b∈ Zsatisfy a| b, then fa | fb.
Discussion of Exercise 4.4.4. Parts (a), (b) and (c) of Exercise 4.4.4 appear (with de-tailed solutions) in [18f-mt1s, Exercise 4]; thus, we shall only sketch the arguments here.
Because of our definition of the fn, the equation
fn = fn−1+ fn−2 (134)
holds for alln∈ Z.
(a)First, prove (by strong induction onn) that
f−n = (−1)n−1 fn for each n∈ N. (135) Then, conclude that the equality f−n = (−1)n−1 fn holds for each n ∈ Z as well, because:
• Ifn is nonnegative, then it follows from (135).
• Ifnis negative, then (135) (applied to−ninstead ofn) yields fn = (−1)−n−1 f−n; but this quickly yields f−n = (−1)n−1 fn.
Thus, Exercise 4.4.4(a)is solved.
An alternative solution for Exercise 4.4.4(a)can be given by strong induction on
|n|. (Note that strong induction on n will not work, since n ranges over Z.) We leave the details to the reader.
(b)First, we claim that
fn+m+1 = fnfm+ fn+1fm+1 for all n ∈Nand m∈ Z. (136) Indeed, (136) can be proved in the same way as we proved (11) for all nonnegative integers n and m (see our “Second attempt at solving Exercise 2.2.3” above) – i.e., by induction onn. (The proof even becomes a bit simpler, since we no longer have to treat the case m = 0 separately: It is no longer problematic that m−1 may be negative.)
we obtain
(−1)n+m fn+m+1 = (−1)n+m(fnfm+ fn+1fm+1).
Dividing both sides of this equality by (−1)n+m, we obtain fn+m+1 = fnfm + fn+1fm+1. This solves Exercise 4.4.4 (b).
An alternative solution to Exercise 4.4.4 (b) can be found in [18f-mt1s, Exercise 4]; it relies on two-sided induction (Theorem 3.1.9).
(c) This is just an exercise in applying the recursive equation of the Fibonacci sequence over and over:77
fn−4+ fn+4
|{z}
(134)
= fn+3+fn+2
= fn−4+ fn+3
|{z}
(134)
= fn+2+fn+1
+ fn+2
|{z}
(134)
= fn+1+fn
= fn−4+fn+2+ fn+1+ fn+1+ fn = fn−4+ fn+2
|{z}
(134)
= fn+1+fn
+2fn+1+fn
= fn−4+fn+1+ fn+2fn+1+ fn = fn−4+2fn+3 fn+1
|{z}
(134)
= fn+fn−1
= fn−4+2fn+3(fn+ fn−1) = fn−4+5fn+3 fn−1
|{z}
(134)
= fn−2+fn−3
= fn−4+5fn+3(fn−2+fn−3) = fn−4+ fn−3
| {z }
(134)
= fn−2
+5fn+3fn−2+2fn−3
= fn−2+5fn+3fn−2+2fn−3 =5fn+2fn−2+2(fn−2+ fn−3)
| {z }
(134)
= fn−1
=5fn+2fn−2+2fn−1=5fn+2(fn−1+ fn−2)
| {z }
(134)
= fn
=5fn+2fn =7fn.
Alternatively, Exercise 4.4.4 (c) can be obtained by applying Exercise 4.4.4 (b) to m=−5 and again tom =3, and then adding the resulting equalities together. (See [18f-mt1s, Exercise 4] for the details.)
(d)Let a,b ∈Zsatisfy a |b. We must prove that fa | fb.
The numbers |a| and |b| belong to N (since a and b belong to Z). Furthermore, Proposition 3.1.3(a)yields|a| | |b|(sincea |b). Hence, Exercise 3.2.2 (applied to|a| and |b| instead ofaand b) yields f|a| | f|b|.
It remains to derive fa | fb from this. But this is easy: Exercise 3.2.2(a)(applied to n = a) yields f−a = (−1)a−1 fa = ±fa. Thus, f|a| =±fa 78. Likewise, f|b| =±fb;
77The symbol “(134)= ” designates an equality that follows from (134). For example, f5(134)= f4+ f3.
78Proof.We are in one of the following two cases:
thus, fb = ±f|b|. Hence, f|b| | fb. Also, from f|a| = ±fa, we obtain fa | f|a|. Thus, fa | f|a| | f|b| | fb. This solves Exercise 4.4.4(d).