The Art and Craft of Problem Solving

9.3 DIFFERENTIATION AND INTEGRATION 337

If k ranges from 1 to n, then k^2 / n^2 ranges from (^1) / n^2 to 1, which suggests that we
should consider a = O,b = 1, and f(x) = 1/(x^2 + 1). It is easy to verify that this
works; i.e.,
� t _
n

= � (f(�) + f(�) + ... + f (�)).

n t:l k^2 + n^2 n n n n

Thus the limit as n ---+ 00 is equal to

r 1 rl dx ]^1 1r 1r

io f(x)dx = io (^1) +x 2 = arctanx 0 =
(^4) - 0 = 4· •
Even finite sums can be analyzed with integrals. If the functions involved are
monotonic, then it is possible to relate integrals and sums with inequalities, as in the
next example.
Example 9.3.9 (Putnam 1996) Show that for every positive integer n,
(
2n
;
1
) 2";1 <^1.^3 .S ... (2n-l) < (
2n: 1
) ¥
Solution: The es in the denominator along with the ugly exponents strongly sug­
gest a simplifying strategy: take logarithms! This transforms the alleged inequality
into
2n-l 2n+l

• 2
  
  
  • (log(2n-l) -1) < S < -
    2
    
    
    • (log(2n+ 1) - 1),
    
    
    
    
  
  
  
  

where

S = log 1 + log3 +logS + ... + log( 2 n - 1).

Let us take a closer look at S. Because logx is monotonically increasing, it is apparent
from the picture (here L1 = 2) that

r2n-1 r2n+1

il

logxdx < 2S <

il

logxdx.

�""-r-1 Y = log x

(^3) 5 7 2n- (^1) 2n+l
The inequality now follows, since an easy application of integration by parts yields
flogxdx =xlogx - x+c. •