lec18 - Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon...

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Unformatted text preview: Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon () Math 16b Autumn 2010 1 / 23 In nite series as areas If the sequence a 1 , a 2 , a 3 ,... consists only of positive numbers, then we could regard the sum i = 1 a i as the area of a region consisting of a rectangle of height a 1 and width one, a rectangle of height a 2 and width one, a rectangle of height a 3 and width one, and so on. Thomas Scanlon () Math 16b Autumn 2010 2 / 23 Integrals and series If f ( x ) is a continuous, decreasing function taking only positive values, then < X n = 2 f ( n ) < Z 1 f ( x ) dx < X n = 1 f ( n ) Consequently, the integral R 1 f ( x ) dx converges if and only if the in nite series n = 1 f ( n ) converges. Thomas Scanlon () Math 16b Autumn 2010 3 / 23 Harmonic series, reconsidered The function f ( x ) = 1 x is continuous, decreasing and only takes positive values for x in the range 1 to . Hence, n = 1 1 n = n = 1 f ( n ) converges if and only if R 1 1 x dx converges. Z 1 1 x dx = lim r Z r 1 1 x dx = lim r ln ( x ) | x = r x = 1 = lim r ln ( r )- ln ( 1 ) = lim r ln ( r ) which diverges. Hence, n = 1 1 n diverges . Thomas Scanlon () Math 16b Autumn 2010 4 / 23 ( 2 ) , again The function f ( x ) = 1 x 2 is continuous, decreasing and only takes positive values for x in the range 1 to . Hence n = 1 1 n 2 = n = 1 f ( n ) converges if and only if R 1 1 x 2 dx converges. Z 1 1 x 2 dx = lim r Z r 1 1 x 2 dx = lim r - 1 x | x = r x = 1 = lim r - 1 r-- 1 1 = 1 Thus, n = 1 1 n 2 converges ....
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lec18 - Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon...

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