Imc2020 day1&2 problems&solutions

IMC 2020 Online
Day 1, July 26, 2020
Problem 1. Let n be a positive integer. Compute the number of words w (nite sequences of
letters) that satisfy all the following three properties:
(1) w consists of n letters, all of them are from the alphabet {a, b, c, d};
(2) w contains an even number of letters a;
(3) w contains an even number of letters b.
(For example, for n = 2 there are 6 such words: aa, bb, cc, dd, cd and dc.)
Armend Sh. Shabani, University of Prishtina
Solution 1. Let N = {1, 2, . . . , n}. Consider a word w that satises the conditions and let
A, B, C, D ⊂ N be the sets of positions of letters a, b, c and d in w, respectively. By the
denition of the words we have A B C D = N. The sets A and B are constrained to have
even sizes.
In order to construct all suitable words w, choose the set S = A∪B rst; by the conditions,
|S| = |A| + |B| must be even. It is well-known that an n-element set (with n ≥ 1) has 2n−1
even subsets, so there are 2n−1
possibilities for S.
If S = ∅ then we can choose C ⊂ N arbitrarily, and then the set D = S C is determined
D uniquely. Since N has 2n
subsets, we have 2n
options for set C and therefore 2n
suitable
words w with S = ∅.
Otherwise, if k = |S| 0, we have to choose an arbitrary subset C of N S and an even
subset A of S; then D = (N S) C and B = S A are determined and |B| = |S| − |A| will
automatically be even. We have 2n−k
choices for C and 2k−1
independent choices for A; so for
each nonempty even S we have 2n−k
· 2k−1
= 2n−1
suitable words.
The number of nonempty even sets S is 2n−1
−1, so in total, the number of words satisfying
the conditions is
1 · 2n
+ (2n−1
− 1) · 2n−1
= 4n−1
+ 2n−1
.
Solution 2. Let an denote the number of words of length n over A = {a, b, c, d} such
that a and b appear even number of times. Further, we dene the following sequences for the
number of words of length n, all over A.
• bn - the number of words with an odd number of a's and even number of b's
• cn - the number of words with even number of a's and an odd number of b's
• dn - the number of words with an odd number of a's and an odd number of b's
We will call them A-words, B-words, C-words and D-words, respectively.
It is clear that a1 = 2 and that
an + bn + cn + dn = 4n
.
First, we nd a recurrence relation for an. If an A-word of length n begins with c or d, it can
be followed by any A-word of length n − 1, contributing with 2an−1. If an A-word of length
n begins with a, it can be followed by any word of length n − 1 that contains an odd number
1

of a's and even number of b's, thus contributing with bn−1. If an A-word of length n begins
with b, it can be followed by any word of length n − 1 that contains even number of a's and
an odd number of b's, thus contributing with cn−1. Therefore we have the following recurrence
relation:
an = 2an−1 + bn−1 + cn−1. (1)
Next, we nd a recurrence relation for bn.
If a B-word of length n begins with c or d, it can be followed by any B-word of length n − 1,
contributing with 2bn−1. If a B-word of length n begins with a, it can be followed by any word of
length n−1 that contains even number of a's and even number of b's, contributing with an−1. If a
B-word of length n begins with b, it can be followed by any word of length n−1 that contains an
odd number of a's and an odd number of b's, contributing with dn−1 = 4n−1
−an−1 −bn−1 −cn−1.
Therefore we have the following recurrence relation:
bn = bn−1 + 4n−1
− cn−1. (2)
Now observe that bk = ck for all k, since simultaneously replacing a's to b's and vice versa
we get a C-word from a B-word. Therefore (2) yields bn = 4n−1
. Now (1) yields
an = 2 · an−1 + 2 · 4n−2
.
Solving the last recurrence relation (for example, diving by 2n
we get xn := an2−n
satises
xn − xn−1 = 2n−3
, and it remains to sum up consecutive powers of 2) we get
an = 2n−1
+ 4n−1
.
Solution 3. Consider the sum
(a + b + c + d)n
+ (−a − b + c + d)n
+ (−a + b + c + d)n
+ (a − b + c + d)n
4
. (∗)
Expanding the parentheses as
(a + b + c + d)n
= (a + b + c + d)(a + b + c + d) . . . (a + b + c + d),
we get a sum of products x1 . . . xn, xi ∈ {a, b, c, d}, naturally corresponding to the words of
length n over the alphabet {a, b, c, d}. Consider the other terms in the numerator similarly.
If a word x1 . . . xn contains A, B, C, D letters a,b,c and d respectively, we get aA
bB
cC
dD
with the coecient
1 + (−1)A+B
+ (−1)A
+ (−1)B
4
=
(1 + (−1)A
)(1 + (−1)B
)
4
=
1, if A and B are even
0, otherwise.
Hence, by substituting a = b = c = d = 1 in (∗) we get the answer (4n
+2n+1
)/4 = 4n−1
+2n−1
.
2

Problem 2. Let A and B be n × n real matrices such that
rk(AB − BA + I) = 1
where I is the n × n identity matrix.
Prove that
trace(ABAB) − trace(A2
B2
) =
1
2
n(n − 1).
(rk(M) denotes the rank of matrix M, i.e., the maximum number of linearly independent
columns in M. trace(M) denotes the trace of M, that is the sum of diagonal elements in M.)
Rustam Turdibaev, V. I. Romanovskiy Institute of Mathematics
Solution. Let X = AB − BA. The rst important observation is that
trace(X2
) = trace(ABAB − ABBA − BAAB + BABA) = 2trace(ABAB) − 2trace(A2
B2
)
using that the trace is cyclic. So we need to prove that trace(X2
) = n(n − 1).
By assumption, X + I has rank one, so we can write X + I = vt
w for two vectors v, w. So
X2
= (vt
w − I)2
= I − 2vt
w + vt
wvt
w = I + (wvt
− 2)vt
w.
Now by denition of X we have trace(X) = 0 and hence wvt
= trace(wvt
) = trace(vt
w) = n so
that indeed
trace(X2
) = n + (n − 2)n = n(n − 1).
An alternative way to use the rank one condition is via eigenvalues: Since X + I has rank
one, it has eigenvalue 0 with multiplicity n−1. So X has eigenvalue −1 with multiplicity n−1.
Since trace(X) = 0 the remaining eigenvalue of X must be n − 1. Hence
trace(X2
) = (n − 1)2
+ (n − 1) · 12
= n(n − 1).
3

Problem 3. Let d ≥ 2 be an integer. Prove that there exists a constant C(d) such that the
following holds: For any convex polytope K ⊂ Rd
, which is symmetric about the origin, and
any ε ∈ (0, 1), there exists a convex polytope L ⊂ Rd
with at most C(d)ε1−d
vertices such that
(1 − ε)K ⊆ L ⊆ K.
(For a real α, a set T ⊂ Rd
with nonempty interior is a convex polytope with at most α vertices, if
T is a convex hull of a set X ⊂ Rd
of at most α points, i.e., T = { x∈X txx | tx ≥ 0, x∈X tx =
1}. For a real λ, put λK = {λx | x ∈ K}. A set T ⊂ Rd
is symmetric about the origin if
(−1)T = T.)
Fedor Petrov, St. Petersburg State University
Solution [in elementary terms] Let {p1, . . . , pm} be an inclusion-maximal collection of points
on the boundary ∂K of K such that the homothetic copies Ki := pi + ε
2
K have disjoint interiors.
We claim that the convex hull L := conv{p1, . . . , pm} satises all the conditions.
First, note that by convexity of K we have aK + bK = (a + b)K for a, b 0. It follows that
Ki ⊂ (1 + ε
2
)K. On the other hand, if k ∈ K, a 0 and and ak ∈ Ki, then
pi ∈ ak −
ε
2
K = ak +
ε
2
K ⊂ (a +
ε
2
)K,
and since pi is a boundary point of K, we get a + ε
2
1, a 1 − ε
2
. It means that all Ki lie
between (1 − ε
2
)K and (1 + ε
2
)K. Since their interiors are disjoint, by the volume counting we
obtain
m
ε
2
d
1 +
ε
2
d
− 1 −
ε
2
d
(3/2)d
ε
(since F(ε) = (1 + ε
2
)d
− (1 − ε
2
)d
is a polynomial in ε without constant term with non-negative
coecients which sum up to (3/2)d
− (1/2)d
), therefore m 3d
ε1−d
.
It is clear that L ⊆ K, so it remains to prove that (1 − ε)K ⊆ L. Assume the contrary:
there exists a point p ∈ (1 − ε)K L. Separate p from L by a hyperplane: Choose a linear
functional such that (p) maxx∈L (x) = maxi (pi). Choose x ∈ K such that (x) =: a is
maximal possible. Note that by our construction x + ε
2
K has a common point with some Ki:
there exists a point z ∈ (x + ε
2
K) ∩ (pi + ε
2
K). We have
(pi) +
ε
2
a (z) (x) −
ε
2
a,
and therefore (pi) a(1 − ε). Since p ∈ (1 − ε)K, we obtain (p) a(1 − ε). A contradiction.
Solution [in the language of Banach spaces] Equip Rd
with the norm · , whose unit
ball is K, call this Banach space V . Choose an inclusion maximal set X ⊂ ∂K whose pairwise
distances are ≥ ε. Put L = convX.
The inclusion L ⊆ K follows from the convexity of K. If the inclusion (1−ε)K ⊆ L fails then
the HahnBanach theorem provides a unit linear functional λ ∈ V ∗
such that max{λ(L)} =
max{λX} ≤ 1 − ε. Then the point x ∈ K, where the maximum max{λ(K)} = 1 is attained
(thanks to the nite dimension and compactness) is in ∂K and, as λ witnesses, at distance ≥ ε
from all other points of L and X, contradicting the inclusion-maximality of X.
The upper bound for the cardinality |X| is obtained by noting that the ε/2 balls centered
at the points of X are pairwise disjoint and lie in the dierence of balls (1+ε/2)K (1−ε/2)K,
whose volume is (1 + ε/2)d
− (1 − ε/2)d
volK, the volume of each of the small balls being
εd
/2d
volK. Hence
|X| ≤
(2 + ε)d
− (2 − ε)d
εd
= O(ε1−d
).
4

Problem 4. A polynomial p with real coecients satises the equation p(x + 1) − p(x) = x100
for all x ∈ R. Prove that p(1 − t) p(t) for 0 t 1/2.
Daniil Klyuev, St. Petersburg State University
Solution 1. Denote h(z) = p(1 − ¯z) − p(z) for complex z. For t ∈ R we have h(it) =
p(1 + it) − p(it) = t100
, h(1/2 + it) = 0.
If p(z) = cnzn
+ . . . + c0, cn = 0, we have
h(a + it) = p((1 − a) + it) − p(a + it) = (1 − 2a) ncnin−1
tn−1
+ Q(t, a)
for some polynomial Q having degree at most n−2 with respect to the variable t. Substituting
a = 0 we get n = 101, cn = 1/101.
Next, for large |t| we see that (h(a + it)) 0 for 0 a 1/2.
Therefore by Maximum Principle for the harmonic function h and the rectangle [0, 1/2] ×
[−N, N] for large enough N we conclude that h is non-negative in this rectangle, in particular
on [0, 1/2], as we need.
Solution 2. Let p(x) = m
j=0 ajxj
. Then
p(x+1)−p(x) =
m
j=0
aj (x + 1)j
− xj
= a1+a2(2x+1)+· · ·+am mxm−1
+
m
2
xm−2
+ · · · + 1 .
This implies that m = 101, mam = 1 so a101 = 1
101
, (m − 1)am−1 + am
m
2
= 0 so a100 = −1
2
etc. For j 1 aj is uniquely dened, a0 may be chosen arbitrarily.
The equality p2n(1
2
) = 0 holds because 0 = p2n(1
2
) + p2n(1 − 1
2
) = 2p2n(1
2
). Let n 1 be an
integer and let pn be a polynomial such that pn(x + 1) − pn(x) = xn
for all x and pn(0) = 0 =
pn(1). The above considerations prove the uniqueness of pn. We have p1(x) = 1
2
x2
− 1
2
x. Also
pn(x + 1) − pn(x) = nxn−1
= n (pn−1(x + 1) − pn−1(x)). Therefore pn(x) = npn−1(x) + cn−1 for
a properly chosen constant cn−1. We shall prove that
(1) p2n−1(x)−p2n−1(1−x) = 0, p2n(x)+p2n(1−x) = 0, c2n = 0, p2n(x) = 2n(2n−1)p2n−2(x)
for n = 1, 2, . . . and for all x. Simple computation shows that p1(x) − p1(1 − x) = 0. We have
(p2(x) + p2(1 − x)) = 2p1(x) + c1 − (2p1(1 − x) + c1) = 0 so the map x → p2(x) + p2(1 − x) is
constant thus p2(x) + p2(1 − x) = p2(0) + p2(1 − 0) = 0. If the rst two equalities hold for some
n then (p2n+1(x) − p2n+1(1 − x)) = (2n + 1)p2n(x) + c2n + (p2n(1 − x) + c2n) = 2c2n so there
exists b ∈ R such that p2n+1(x) − p2n+1(1 − x) = 2c2nx + b for all x. p2n+1(0) − p2n+1(1 − 0) = 0
and p2n+1(1)−p2n+1(1−1) = 0 so 2c2n = 0 = b. This proves that p2n+1(x)−p2n+1(1−x) = 0 for
all x. In a similar way we shall prove the second equality: (p2n+2(x) + p2n+2(1 − x)) = (2n +
2)p2n+1(x)+c2n+1 −(2n+2) (p2n+1(1 − x) + c2n+1) = 0 so the map x → p2n+2(x)+p2n+2(1−x)
is constant hence p2n+2(x) + p2n+2(1 − x) = p2n+2(0) + p2n+2(1 − 0) = 0 for all x. Now
p2n+2(x) = ((2n + 2)p2n+1(x) + c2n+1) = (2n + 2)p2n+1(x) = (2n + 2)((2n + 1)p2n(x) + c2n) =
(2n+2)(2n+1)p2n(x). Since p2(x) = 2p1(x)+c1 = x2
−x+c1 we obtain p2(x) = 2x−1 0 for
x 1
2
.The function p2 is strictly concave on [0, 1
2
] and p2(0) = 0 = p2(1
2
). Therefore p2(x) 0
for x ∈ (0, 1
2
). This together with the equality p4(x) = 12p2(x) implies that p4 is strictly convex
on [0, 1
2
] so in view of p4(0) = 0 = p4(1
2
) we conclude that p4(x) 0 for x ∈ (0, 1
2
). Easy
induction shows that for x ∈ (0, 1
2
) one has p2n(x) 0 for an odd n and p2n(x) 0 for an even
n. If t ∈ (0, 1
2
) then by (1) we get p100(1 − t) − p100(t) = −2p100(t) 0 as required.
5

IMC 2020 Online
Day 2, July 27, 2020
Problem 5. Find all twice continuously dierentiable functions f : R → (0, +∞) satisfying
f (x)f(x) ≥ 2(f (x))2
for all x ∈ R.
Karen Keryan, Yerevan State University American University of Armenia, Yerevan
Solution. We shall show that only positive constant functions satisfy the condition.
Let g(x) =
1
f(x)
. Notice that
g =
1
f
=
−f
f2
=
2(f )2
− f f
f3
≤ 0,
so the positive function g(x) is concave. We show that g must be constant.
Take two arbitrary real numbers a b. By the concavity of g, for all u a and v b we
have
g(a) − g(u)
a − u
≥
g(b) − g(a)
b − a
≥
g(v) − g(b)
v − b
.
Combining this with g(u), g(v) 0 we get
g(a)
a − u

g(b) − g(a)
b − a

−g(b)
v − b
Now by taking limits u → −∞ and v → ∞ we obtain
0 ≥
g(b) − g(a)
b − a
≥ 0,
so g(a) = g(b). This holds for any pair (a, b), so g(x) is constant and f(x) = 1/g(x) also is
constant.
If f is constant then f = f = 0, so the condition is satised.
Remark. Instead of the function 1/f(x), the same idea works with arctan f(x):
(arctan f(x)) =
f (1 + f2
) − 2(f )2
(1 + f2)2
=
f (1 + f2
) − 2(f )2
(1 + f2
)
(1 + f2)2
=
f − 2(f )2
1 + f2
≥ 0.
As can be seen, arctan f(x) is a bounded convex function, therefore it must be constant.
Problem 6. Find all prime numbers p for which there exists a unique a ∈ {1, 2, . . . , p} such
that a3
− 3a + 1 is divisible by p.
Géza Kós, Loránd Eötvös University, Budapest
Solution 1. We show that p = 3 the only prime that satses the condition.
1

Let f(x) = x3
− 3x + 1. As preparation, let's compute the roots of f(x). By Cardano's
formula, it can be seen that the roots are
2Re
3 −1
2
+
−1
2
2
−
−3
3
3
= 2Re
3
cos
2π
3
+ i sin
2π
3
= 2 cos
2π
9
, 2 cos
4π
9
, 2 cos
8π
9
where all three values of the complex cubic root were taken.
Notice that, by the trigonometric identity 2 cos 2t = (2 cos t)2
− 2, the map ϕ(x) = x2
− 2
cyclically permutes the three roots. We will use this map to nd another root of f, when it is
considered over Fp.
Suppose that f(a) = 0 for some a ∈ Fp and consider
g(x) =
f(x)
x − a
=
f(x) − f(a)
x − a
= x2
+ ax + (a2
− 3).
We claim that b = a2
− 2 is a root of g(x). Indeed,
g(b) = (a2
− 2)2
+ a(a2
− 2) + (a2
− 3) = (a + 1) · f(a) = 0.
By Vieta's formulas, the other root of g(x) is c = −a − b = −a2
− a + 2.
If f has a single root then the three roots must coincide, so
a = a2
− 2 = −a2
− a + 2.
Here the quadratic equation a = a2
− 2, or equivalently (a + 1)(a − 2) = 0, has two solutions,
a = −1 and a = 2. By f(−1) = f(2) = 3, in both cases we have 0 = f(a) = 3, so the only
choice is p = 3.
Finally, for p = 3 we have f(1) = −1, f(2) = 3 and f(3) = 19, from these values only f(2)
is divisible by 3, so p = 3 satises the condition.
Solution 2 (outline) Dene f(x) and g(x) like in Solution 1. The discriminant of g(x) is
∆g = a2
− 4(a2
− 3) = 12 − 3a2
.
We show that ∆g has a square root in Fp.
Take two integers k, m (to be determinated later) and consider
∆g = ∆g + (ka + m)f(a) = ka4
+ ma3
− (3k + 1)a2
+ (k − 3m)a + (m + 12).
Our goal is to choose k, m in such a way that the last expression is a complete square. Either
by direct calculations or guessing, we can nd that k = m = 4 works:
∆g = ∆g + (4a + 4)f(a) = 4a4
+ 4a3
− 15a2
− 8a + 16 = (2a2
+ a − 4)2
.
If p = 2 then we can conclude that f(x) has either no or three roots, therefore p is suitable
if and only is f(x) is a complete cube: x3
− 3x + 1 = (x − a)3
. From Vieta's formulas a3
= 1,
so a = 0 and 3a = 0, which is possible if p = 3.
For p = 3 we have f(x) = (x + 1)3
, so p = 3 is suitable.
The case p = 2 must be checked separately because the quadratic formula contains a division
by 2. f(1) = −1 and f(2) = 3, so p = 2 is not suitable.
Solution 3 (outline) Assume p 3; the cases p = 2 and p = 3 will be checked separately.
2

Let f(x) = x3
− 3x + 1 and suppose that a ∈ Fp is a root of f(x), and let b, c ∈ Fp2 be the
other two roots. The discriminant ∆f of f(x) can be expressed by the elementary symmetric
polynomials of a, b, c; it can be calculated that
∆f = (b − c)2
(a − b)2
(a − c)2
= 81 = 92
,
so
(b − c)(a − b)(a − c) = ±9 ∈ Fp.
Notice that ∆f = 0, so the three roots are distinct.
Either b, c ∈ Fp or b, c are conjugate elements in Fp2 Fp, we have (a − b)(a − c) ∈ Fp, so
b − c = (b−c)(a−b)(a−c)
(a−b)(a−c)
∈ Fp. From Vieta's formulas we have b + c ∈ Fp as well; since p = 2, it
follows that b, c ∈ Fp. Now f(x) has three distinct roots in Fp, so p cannot be suitable.
p = 2 does not satises the condition because both f(1) = −1 and f(2) = 3 are odd. p = 3
is suitable, because f(2) = 3 is divisible by 3 while f(1) = −1 and f(3) = 19 are not.
Problem 7. Let G be a group and n ≥ 2 be an integer. Let H1 and H2 be two subgroups of
G that satisfy
[G : H1] = [G : H2] = n and [G : (H1 ∩ H2)] = n(n − 1).
Prove that H1 and H2 are conjugate in G.
(Here [G : H] denotes the index of the subgroup H, i.e. the number of distinct left cosets
xH of H in G. The subgroups H1 and H2 are conjugate if there exists an element g ∈ G such
that g−1
H1g = H2.)
Ilya Bogdanov and Alexander Matushkin, Moscow Institute of Physics and Technology
Solution 1. Denote K = H1 ∩ H2. Since
n(n − 1) = [G : K] = [G : H1][H1 : K] = n[H1 : K],
we obtain that [H1 : K] = n − 1. Thus, the subgroup H1 is partitioned into n − 1 left cosets of
K, say H1 = n−1
i=1 hiK. Therefore, the set H1H2 = {ab: a ∈ H1, b ∈ H2} is partitioned as
H1H2 =
n−1
i=1
hiK H2 =
n−1
i=1
hiKH2 =
n−1
i=1
hiH2.
The last equality holds because K ⊆ H2, so KH2 = H2. The last expression is a disjoint
union since
hiH2 ∩ hjH2 = ∅ ⇐⇒ h−1
i hj ∈ H2 ⇐⇒ h−1
i hj ∈ K ⇐⇒ hi = hj.
Thus, H1H2 is a disjoint union of n−1 left cosets with respect to H2; hence L = G(H1H2)
is the remaining such left coset. Similarly, L is a right coset with respect to H1. Therefore, for
each g ∈ L we have L = gH2 = H1g, which yields H2 = g−1
H1g.
Solution 2. Put G/H1 = X and G/H2 = Y , those are n-element sets acted on by G from the
left. Let G act on X × Y from the left coordinate-wise, consider this product as a table, with
rows being copies of X and columns being copies of Y .
The stabilizer of a point (x, y) in X × Y is H1 ∩ H2. By the orbit-stabilizer theorem, we
obtain that the orbit Z of (x, y) has size [G : H1 ∩ H2] = n(n − 1).
If Z contains a whole column then there is a subgroup G1 of G that stabilizes x and acts
transitively on Y . If we conjugate G1 to a group G1, then its action remains transitive on Y ,
3

so by conjugation we obtain columns of the table. Since G acts transitively on X, we cover all
the columns. It follows that Z = X × Y , so
n(n − 1) = |Z| = |X × Y | = n2
,
which is a contradiction.
Hence every column of X × Y has an element not from Z. The same holds for the rows of
X × Y . There are n elements not from Z in total and they induce a bijection between X and
Y which allows us to identify X = Y .
After this identication, every element (x, x) from the diagonal of X × X (i.e. from (X ×
X) Z) is moved to a diagonal element by any g ∈ G, because gx = gx. In this formula the
action of g in the left hand side and the action of g in the right hand side are the actions of g
on X and Y respectively.
Therefore our bijection between X and Y is an isomorphism of sets with a left action of G.
Since H1 and H2 are stabilizers of the points in the same transitive action of G, we conclude
that they are conjugate.
Remark. The situation in the problem is possible for every n ≥ 2: let G = Sn and let H1
an H2 be the stabilizer subgroups of two elements.
4

Problem 8. Compute
lim
n→∞
1
log log n
n
k=1
(−1)k n
k
log k.
(Here log denotes the natural logarithm.)
Fedor Petrov, St. Petersburg State University
Solution 1. Answer: 1.
The idea is that if f(k) = gk
, then
(−1)k n
k
f(k) = (1 − g)n
.
To relate this to logarithm, we may use the Frullani integrals
∞
0
e−x
− e−kx
x
dx = lim
c→+0
∞
c
e−x
x
dx −
∞
c
e−kx
x
dx = lim
c→+0
∞
c
e−x
x
dx −
∞
kc
e−x
x
dx =
lim
c→+0
kc
c
e−x
x
dx = log k + lim
c→+0
kc
c
e−x
− 1
x
dx = log k.
This gives the integral representation of our sum:
A :=
n
k=1
(−1)k n
k
log k =
∞
0
−e−x
+ 1 − (1 − e−x
)n
x
dx.
Now the problem is reduced to a rather standard integral asymptotics.
We have (1−e−x
)n
1−ne−x
by Bernoulli inequality, thus 0 −e−x
+1−(1−e−x
)n
ne−x
,
and we get
0
∞
M
−e−x
+ 1 − (1 − e−x
)n
x
dx n
∞
M
e−x
x
dx nM−1
∞
M
e−x
dx = nM−1
e−M
.
So choosing M such that MeM
= n (such M exists and goes to ∞ with n) we get
A = O(1) +
M
0
−e−x
+ 1 − (1 − e−x
)n
x
dx.
Note that for 0 x M we have e−x
e−M
= M/n, and (1 − e−x
)n−1
e−e−x(n−1)
e−M(n−1)/n
tends to 0 uniformly in x. Therefore
M
0
(1 − e−x
)(1 − (1 − e−x
)n−1
)
x
dx = (1 + o(1))
M
0
1 − e−x
x
dx.
Finally
M
0
1 − e−x
x
dx =
1
0
1 − e−x
x
dx +
M
1
−e−x
x
dx +
M
1
dx
x
=
log M + O(1) = log(M + log M) + O(1) = log log n + O(1),
and we get A = (1 + o(1)) log log n.
Solution 2. We start with a known identity (a nite dierence of 1/x).
5

Expand the rational function
f(x) =
m!
x(x + 1) . . . (x + m)
as the linear combination of simple fractions f(x) = m
j=0 cj/(x + j). To nd cj we use
cj = ((x + j)f(x)) |x=−j = (−1)j m
j
.
So we get
m
k=0
(−1)k m
k
1
x + k
=
m!
x(x + 1) . . . (x + m)
. (1)
Another known identity we use is
n
k=j+1
(−1)k n
j
=
n
k=j+1
(−1)k n − 1
k
+
n − 1
k − 1
= (−1)j+1 n − 1
j
. (2)
Finally we write log k =
k
1
dx
x
= k−1
j=1 Ij, where Ij =
1
0
dx
x+j
.
Now we have
S :=
n
k=1
(−1)k n
k
log k =
n
k=1
(−1)k n
k
k−1
j=1
Ij =
n−1
j=1
Ij
n
k=j+1
(−1)k n
k
=
(2)
n−1
j=1
Ij(−1)j+1 n − 1
j
=
1
0
n−1
j=1
(−1)j+1 n − 1
j
dx
x + j
=
1
0
1
x
−
n−1
j=0
(−1)j n − 1
j
dx
x + j
dx =
(1)
1
0
1
x
−
(n − 1)!
x(x + 1) . . . (x + (n − 1))
dx =
1
0
dx
x
1 −
1
(1 + x)(1 + x/2) . . . (1 + x/(n − 1))
.
So S is again expressed as an integral, for which it is not hard to get an asymptotics.
Since et
1+t for all real t (by convexity or any other reason), we have ey2−y
1+y2
−y =
1+y3
1+y
1
1+y
and 1
1+y
1
ey = e−y
for y 0. Therefore
ey2−y 1
1 + y
e−y
, y 0.
Using this double inequality we get
e
x2 1+ 1
22 +...+ 1
(n−1)2 −x(1+1
2
+...+ 1
n−1 ) 1
(1 + x)(1 + x/2) . . . (1 + x/(n − 1))
e−x(1+1
2
+...+ 1
n−1 ).
Since x2
(1 + 1/22
+ . . .) 2x2
2x, we conclude that
1
(1 + x)(1 + x/2) . . . (1 + x/(n − 1))
= e−Cnx
, where − 2 +
n−1
j=1
1
j
Cn
n−1
j=1
1
j
,
i.e., Cn = log n + O(1). Thus
S =
1
0
dx
x
(1 − e−Cnx
) =
Cn
0
dt
t
(1 − e−t
) =
Cn
1
dt
t
+
1
0
(1 − e−t
)
dt
t
+
Cn
1
e−t dt
t
= log Cn + O(1) = log log n + O(1).
6

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