542 Real Analysis
and, using the inequalityex> 1 +x,
2 i/n
1 +ni^1= 2 (i−^1 )/n21 /n
1 +ni^1= 2 (i−^1 )/neln 2/n
1 +ni^1> 2 (i−^1 )/n1 +ln 2n
1 +ni^1> 2 (i−^1 )/n,fori≥2. By the intermediate value property, for eachi≥2 there existsξi∈[i−n^1 ,ni]
such that
2 i/n
1 +ni^1= 2 ξi.Of course, the term corresponding toi=1 can be neglected whennis large. Now we see
that our limit is indeed the Riemann sum of the function 2xintegrated over the interval
[ 0 , 1 ]. We obtain
lim
n→∞(
21 /n
n+ 1+
22 /n
n+^12+···+
2 n/n
n+^1 n)
=
∫ 1
02 xdx=1
ln 2.
(Soviet Union University Student Mathematical Olympiad, 1976)471.This is an example of an integral that is determined using Riemann sums. Divide
the interval[ 0 ,π]intonequal parts and consider the Riemann sum
π
n[
ln(
a^2 − 2 acos
π
n+ 1
)
+ln(
a^2 − 2 acos
2 π
n+ 1
)
+···
+ln(
a^2 − 2 acos(n− 1 )π
n+ 1
)]
.
This expression can be written as
π
nln(
a^2 − 2 acos
π
n+ 1
)(
a^2 − 2 acos
2 π
n+ 1
)
···
(
a^2 − 2 acos
(n− 1 )π
n+ 1
)
.
The product inside the natural logarithm factors as
n∏− 1k= 1[
a−(
coskπ
n
+isinkπ
n)][
a−(
coskπ
n
−isinkπ
n)]
.
These are exactly the factors ina^2 n−1, except fora−1 anda+1. The Riemann sum
is therefore equal to
π
nln
a^2 n− 1
a^2 − 1