492 Real Analysis
0 =f 1 ( 0 )<x 2 =f 1 (x 1 )=x 1 −x^21 ≤1
4
<
2
4
,
0 =f 2 ( 0 )<x 3 =f 2 (x 2 )=x 2 − 2 x 22 ≤1
8
<
2
9
,
0 =f 3 ( 0 )<x 4 =f 3 (x 3 )=x 3 − 3 x 32 ≤1
12
<
2
16
.
Here we used the inequalityx 1 −x 12 −^14 =−(x 1 −^12 )^2 ≤0 and the like. Now assume that
the inequality is true forn≥4 and prove it forn+1. Sincen≥, we havexn≤n^22 ≤ 21 n.
Therefore,
0 =fn( 0 )<xn+ 1 =fn(xn)≤fn(
2
n^2)
=
2
n^2−n·4
n^4=
2 n− 4
n^3.
It is an easy exercise to check that
2 n− 4
n^3<
2
(n+ 1 )^2,
which then completes the induction.
We conclude that the series
∑
nxnhas positive terms and is bounded from above by
the convergentp-series 2
∑
n1
n^2 , so it is itself convergent.
(Gazeta Matematic ̆a (Mathematics Gazette, Bucharest), 1980, proposed by L.
Panaitopol)
360.The series is convergent because it is bounded from above by the geometric series
with ratio^12. Assume that its sum is a rational numberab. Choosensuch thatb< 2 n.
Then
a
b−
∑nk= 11
2 k^2=
∑
k≥n+ 11
2 k^2.
But the sum
∑n
k= 1
1
2 k^2 is equal tom
2 n^2 for some integern. Hencea
b−
∑nk= 11
2 k^2=
a
b−
m
2 n^2>
1
2 n^2 b>
1
2 n^2 +n>
1
2 (n+^1 )^2 −^1=
∑
k≥(n+ 1 )^21
2 k>
∑
k≥n+ 11
2 k^2,
a contradiction. This shows that the sum of the series is an irrational number.
Remark.In fact, this number is transcendental.
361.The series is bounded from above by the geometric series|a 0 |( 1 +|z|+|z|^2 +···),
so it converges absolutely. Using the discrete version of integration by parts, known as
the Abel summation formula, we can write