Limits and Derivatives · Question 18

Let [t] be the greatest integer less than or equal to t. Then the least value of p N for which x 0+ (x ( [ 1x ] + [ 2x ] + + [ px ] ) - x2 ( [ 1x2 ] + [ 22x2 ] + + [ 92x2 ] ) ) 1 is equal to .

Answer

24

Solution

[ kx ] = kx - \ kx \, where 0 \ \ < 1

x [ kx ] = k - x \ kx \ k as x 0+

Similarly, x2 [ k2x2 ] = k2 - x2 \ k2x2 \ k2

limit = (1 + 2 + + p) - (12 + 22 + + 92)

= p(p + 1)2 - 9 10 196 = p(p + 1)2 - 285

Condition: p(p + 1)2 - 285 1 p(p + 1) 572

p = 23: 23 24 = 552 < 572; p = 24: 24 25 = 600 572

least value of p = 24

Answer: 24