Advanced book on Mathematics Olympiad

Real Analysis 541

468.We have

sn=

1

√

4 n^2 − 12

+

1

√

4 n^2 − 22

+···+

1

√

4 n^2 −n^2

=

1

n

⎡

⎣√^1

4 −

( 1

n

) 2 +

1

√

4 −

( 2

n

) 2 +···+

1

√

4 −

(n

n

) 2

⎤

⎦.

Hencesnis the Riemann sum of the functionf:[ 0 , 1 ]→R,f(x)=√^1
4 −x^2
associated

to the subdivisionx 0 = 0 <x 1 =^1 n <x 2 =^2 n <···<xn = nn =1, with the
intermediate pointsξi=ni ∈[xi,xi+ 1 ]. The answer to the problem is therefore

lim

n→∞

sn=

∫ 1

0

1

√

4 −x^2

dx=arcsin

x

2

∣∣

∣

1

0

=

π

6

.

469.Write the inequality as

1

n

∑n

i= 1

1

√

(^2) ni+ 5

<

√

7 −

√

5.

The left-hand side is the Riemann sum of the strictly decreasing functionf(x)=√ 21 x+ 5.
This Riemann sum is computed at the right ends of the intervals of the subdivision of
[ 0 , 1 ]by the pointsni,i= 1 , 2 ,...,n−1. It follows that

1

n

∑n

i= 1

1

√

2 in+ 5

<

∫ 1

0

1

√

2 x+ 5

dx=

√

2 x+ 5

∣

∣∣

∣

1

0

=

√

7 −

√

5 ,

the desired inequality.
(communicated by E. Craina)

470.We would like to recognize the general term of the sequence as being a Riemann
sum. This, however, does not seem to happen, since we can only write

∑n

i= 1

2 i/n

n+^1 i

=

1

n

∑n

i= 1

2 i/n

1 +ni^1

.

But fori≥2,

2 i/n>

2 i/n

1 +ni^1

,