Advanced book on Mathematics Olympiad

490 Real Analysis ∑ xj< 10 n 1 xj = ∑n i= 1 ∑ 10 i−^1 ≤xj< 10 i 1 xj < ∑n i= 1 ∑ 10 i−^1 ≤xj< 10 i 1 10 i−^1 = ∑n i= ...

Real Analysis 491 let bnk+ 1 = ··· =bnk+ 2 rk= 1 2 rk and ank+ 1 = ··· =ank+ 2 rk= 1 2 k· 2 rk . Ifkis odd, we do precisely the ...

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 ...

Real Analysis 493 a 0 +a 1 z+a 2 z^2 +···+anzn+··· =(a 0 −a 1 )+(a 1 −a 2 )( 1 +z)+···+(an−an+ 1 )( 1 +z+···+zn)+···. Assume tha ...

494 Real Analysis The problem now repeats forw 1 , which is irrational and between 0 and 1. Againp 1 has to be the unique intege ...

Real Analysis 495 ∑ m∈M ∑∞ k= 2 ∑∞ j= 1 1 mkj = ∑ m∈M ∑∞ j= 1 ∑∞ k= 2 1 mkj . Again, we can change the order of summation becaus ...

496 Real Analysis Therefore, ∑ p≤n 1 p + ∑ p≤n 1 p^2 >ln lnn. But ∑ p≤n 1 p^2 < ∑∞ n= 2 1 k^2 = π^2 6 − 1 < 1. Hence ∑ ...

Real Analysis 497 = m∑− 1 n= 0 [ ( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn)−( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn+^1 ) ] = 1 − 0 = 1 , and the identity ...

498 Real Analysis 12 − 22 + 32 −···+(− 1 )l+^2 (l+ 1 )^2 =(− 1 )l+^1 l(l+ 1 ) 2 +(− 1 )l+^2 (l+ 1 )^2 =(− 1 )l+^2 (l+ 1 ) ( − l ...

Real Analysis 499 = 1 4 (√ n^2 +(n+ 1 )^2 − √ n^2 +(n− 1 )^2 ) = 1 4 (bn+ 1 −bn), wherebn= √ n^2 +(n− 1 )^2. Hence the given sum ...

500 Real Analysis 374.First solution: LetSn= ∑n k= 0 (−^1 ) k(n−k)!(n+k)!.Reordering the terms of the sum, we have Sn=(− 1 )n ∑n ...

Real Analysis 501 This telescopes to 1 2 (n+ 1 ) [n!(n+ 1 )!+(− 1 )n( 2 n+ 1 )!]. (T. Andreescu, second solution by R. Stong) 37 ...

502 Real Analysis 377.ForN≥2, define aN= ( 1 − 4 1 )( 1 − 4 9 )( 1 − 4 25 ) ··· ( 1 − 4 ( 2 N− 1 )^2 ) . The problem asks us to ...

Real Analysis 503 Hence 1 Pn+ 1 − 1 Pn = xn+ 2 (n+ 1 )! − xn+ 1 n! = xn+ 2 −(n+ 1 )xn+ 1 (n+ 1 )! = xn+^1 (n+ 1 )! , forn≥ 1. Ad ...

504 Real Analysis =exp ( cosu− 1 u · u sinu ) =e^0 ·^1 =e^0 = 1. The limit therefore exists. 382.Without loss of generality, we ...

Real Analysis 505 whose limit astgoes to infinity is 1. The squeezing principle implies that tlim→∞ f (mt) f(t) = 1 , as desired ...

506 Real Analysis ... Figure 66 3 f (m 3 n ) = 2 f (m− 1 3 3 n−^1 ) +f (m+ 2 3 3 n−^1 ) = 0 + 0 = 0. Thusf( 3 mn)=0. Finally, if ...

Real Analysis 507 g(c+ 2 h 0 )>g(c+h 0 )>g ( c+ h 0 2 ) >···>g ( c+ h 0 2 n ) >···. Passing to the limit, we obta ...

508 Real Analysis ma =f( 0 ), and hencefis constant on[ 0 ,a]. Passing to the limit witha→1, we conclude thatfis constant on the ...

Real Analysis 509 differ fromf(x)by at most 2 m^1 + 1. We conclude again that asn→∞,f(xn)→f(x). This proves the continuity off. ...