Advanced book on Mathematics Olympiad

Real Analysis 469

P(k)≈

μk

ek·k!

.

The exercise we just solved shows that this approximation is good.

315.Let us assume that the answer is negative. Then the sequence has a bounded
subsequence(xnk)k. The set{xxnk |k∈Z}is finite, since the indicesxnkbelong to a
finite set. Butxxnk =n^4 k, and this takes infinitely many values fork≥1. We reached
a contradiction that shows that our assumption was false. So the answer to the question
is yes.
(Romanian Mathematical Olympiad, 1978, proposed by S. Radulescu) ̆

316.Define the sequence(bn)nby

bn=max{|ak|, 2 n−^1 ≤k< 2 n}.

From the hypothesis it follows thatbn≤ bn 2 −^1. Hence 0≤bn≤ 2 bn−^11 , which implies
that(bn)nconverges to 0. We also have that|an|≤bn, forn≥1, so by applying the
squeezing principle, we obtain that(an)nconverges to zero, as desired.
(Romanian Mathematical Olympiad, 1975, proposed by R. Gologan)

317.First solution: Using the fact that limn→∞n

√

a =1, we pass to the limit in the
relation from the statement to obtain

(^1) ︸+ 1 +···+︷︷ (^1) ︸
ktimes
= (^1) ︸+ 1 +···+︷︷ (^1) ︸
mtimes

.

Hencek=m. Using L’Hôpital’s theorem, one can prove that limx→ 0 x(ax− 1 )=lna,
and hence limn→∞n(n

√

a− 1 )=lna. Transform the relation from the hypothesis into

n(n

√

a 1 − 1 )+···+n(n

√

ak− 1 )=n(n

√

b 1 − 1 )+···+n(n

√

bk− 1 ).

Passing to the limit withn→∞, we obtain

lna 1 +lna 2 +···+lnak=lnb 1 +lnb 2 +···+lnbk.

This implies thata 1 a 2 ···ak=b 1 b 2 ···bk, and we are done.

Second solution: FixN>k; then takingn=(Nm!)for 1≤m≤k, we see that the power-
sum symmetric polynomials inai^1 /N!agree with the power-sum symmetric polynomials
inbi^1 /N!. Hence the elementary symmetric polynomials in these variables also agree and
hence there is a permutationπsuch thatbi=aπ(i).
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
D. Andrica, second solution by R. Stong)