474 Real Analysis
···
1
n<lnn−ln(n− 1 ),we obtain
1 +1
2
+
1
3
+···+
1
n
< 1 +lnn< 1 +ln(n+ 1 ).Therefore,an<1, for alln. We found that the sequence is increasing and bounded,
hence convergent.
324.The sequence is increasing, so all we need to show is that it is bounded. The main
trick is to factor a
√
- The general term of the sequence becomes
an=√
2
√√
√√
√^1
2
+
√√
√
√^2
4
+
√
3
8
+···+
√
n
2 n<
√
2
√
1 +
√
1 +
√
1 +···+
√
1.
Letbn=
√
1 +
√
1 +···+
√
1, where there arenradicals. Thenbn+ 1 =√
1 +bn.We
see thatb 1 = 1 <2, and ifbn<2, thenbn+ 1 <
√
1 + 2 <2. Inductively we prove
thatbn<2 for alln. Therefore,an< 2
√
2 for alln. Being monotonic and bounded, the
sequence(an)nis convergent.
(Matematika v škole, 1971, solution from R. Honsberger, More Mathematical
Morsels, Mathematical Association of America, 1991)
325.We examine first the expression under the square root. Its zeros are−^1 ±
√
5
2. In order
for the square root to make sense,anshould be outside the interval(−^1 −
√ 5
2 ,− 1 +√ 5
2 ).
Sincean≥0 forn≥2, being the square root of an integer, we must havean≥−^1 +
√ 5
2
forn≥2. To simplify the notation, letr=−^1 +
√
5
2.
Now suppose by contradiction thata 1 ∈(− 2 , 1 ). Thena 22 =a^21 +a 1 − 1 =(
a 1 +1
2
) 2
−
5
4
<
(
3
2
) 2
−
5
4
= 1 ,
soa 2 ∈[r, 1 ). Now ifan∈[r, 1 ), then
an^2 + 1 =a^2 n+an− 1 <a^2 n< 1.Inductively we prove thatan∈[r, 1 )andan+ 1 <an. The sequence(an)nis bounded
and strictly decreasing; hence it has a limitL. This limit must lie in the interval[r, 1 ).