Assignment 16 Solution. Please do not copy and paste my answer. You will get similar questions but with different numbers!

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1 Assignment 6 Solution Please do not copy and paste my answer. You will get similar questions but with different numbers! Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= only if the improper integral (i). If n= (ii). If n= dx = x 5 4x 4 t dx = ] t x 5 4x 4 = + 4t 4 4 Since the integral is finite, the series is convergent. t 3 x 5 = 3 lim dx = 3 lim t x5 [ t 4t ] = 3 4

2 Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= only if the improper integral (i). If n= (ii). If n= t dx = lim x+5 t dx = lim x+5 t t =. Let u = x + 5, du = dx, then dx = u du = u = x + 5 x+5 x+5 dx = x + 5 t = t Since the integral is not finite, the series diverges.

3 Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= only if the improper integral (i). If n= (ii). If n= lim x x +6 dx = lim t x x +6 dx = lim t ln t + 6 ln7 =. (You can try integration using partial fraction. But I think u-substitution is easier.) Let u = x + 6, then du = xdx or xdx = du. x x +6 dx = u du = ln x + 6. x x +6 dx = ln x + 6 ] t = ln t + 6 ln7 Since the integral is not finite, the series diverges. 3

4 The p-series n= n p is convergent if p > and divergent if p. a = =, a =,... So a n = n n = n 3. Here p = 3 >. It is convergent. 4

5 The p-series n= n p is convergent if p > and divergent if p. (Comparison Test). Suppose that n= a n and n= b n are series with positive terms. (i) If n= b n is convergent and a n b n for all n, then n= a n is also convergent. (ii) If n= b n is divergent and a n b n for all n, then n= a n is also divergent. 7 = 6 +, 3 = 6 +, 9 = 6 3 +,... So a n = 6n+. Since n, 6n + n 6n + 6n+n 6n+. Let b n = 6n+n = 7n. bn diverges because p =. By Comparison test, 5

6 The p-series n= n p is convergent if p > and divergent if p. (Law of Series) If n= a n and n= b n are both convergent, then so are the series n= ca n, n= (a n + b n ) and n= (a n b n ). ca n = c n= n= a n n= (a n + b n ) = a n + n= n= b n n= (a n b n ) = a n n= b n n= n+7 n a n = = + 7 = + 7 n n n n 3 n Then first term p = 3, so converges. n 3 The second term p =, so 7 n Then a n = n 3 converges. + 7 n also converges. 6

7 The p-series n= n p is convergent if p > and divergent if p. (Comparison Test). Suppose that n= a n and n= b n are series with positive terms. (i) If n= b n is convergent and a n b n for all n, then n= a n is also convergent. (ii) If n= b n is divergent and a n b n for all n, then n= a n is also divergent. a n = 7 n +6. n + 6 n n +6 n. So 7 n +6 7 = b n n and b n converges as p=>. So a n also converges. 7

8 The p-series n= n p is convergent if p > and divergent if p. (Comparison Test). Suppose that n= a n and n= b n are series with positive terms. (i) If n= b n is convergent and a n b n for all n, then n= a n is also convergent. (ii) If n= b n is divergent and a n b n for all n, then n= a n is also divergent. Here is an easier way. ln x x for x > 0. So lnn n and lnn n 7 test, lnn n 7 converges. Or you can use the integral test. n n 7 = n 6 n 7. Both of them converges as p = 6 and p = 7. So [ n 6 n 7 ] converges. By the comparison Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= only if the improper integral (i). If n= (ii). If n= f (x) = ln x x 7 is continuous, positive on [, ). f (x) = 7ln x x 8 Let u = ln x, dv = dx, then du = x 7 x dx and v = 6 ln x dx = uv vdu = x 7 6 ln x + x 6 6 Then ln x x 6. x 6 x dx = ln x 6 + x 6 6 ln x x 6 36 ] t x 6 = ln t 6 t t 6 36 dx = x 7 6 ln x t dx = lim x 7 t ln x dx = lim x 7 t ln t 6 t t 6 36 = lim t 6 t t = lim 6t 5 t = 0 by LH rule. 6t 6 lim t ln t So ln x dx = lim x 7 t ln t 6 t t 6 36 = 36. Since the integral converges, the original series converges. < 0. So we can try integral test. dx = ln x x x 6 6 ( 6 ) = ln x x 6 6 x 6 36 x 6 8

9 Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= only if the improper integral (i). If n= (ii). If n= f (x) = x ln x is continuous and positive on [, ). f (x) = +ln x (x ln x) < 0. Let u = ln x, then du = x dx. Then x ln x dx = u du = ln u = ln ln x. x ln x dx = lim t x ln x dx = lim t [ln ln t ln ln ] =. So the original series also diverges. 9

10 Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= only if the improper integral (i). If n= (ii). If n= f (x) = x is positive and continuous on [, ). f (x) = x( 5x) < 0 if x < 0 or x > e 5x e 5x 5. So f (x) < 0 on [, ). Let u = x and dv = e 5x dx, then du = xdx and v = 5 e 5x. x dx = e 5x 5 x e 5x + 5 xe 5x dx. Apply Integration by parts again for xe 5x dx. Let u = x and dv = e 5x dx, then du = dx and v = 5 e 5x. xe 5x dx = uv vdu = 5 xe 5x + 5 e 5x dx = 5 xe 5x 5 e 5x. Now we have x dx = e 5x 5 x e 5x + 5 xe 5x dx = 5 x e 5x + 5 [ 5 xe 5x 5 e 5x ] = 5 x e 5x 5 xe 5x 5 e 5x lim t x dx = lim e 5x t 5 x e 5x 5 xe 5x 5 e 5x ] t = lim t 5 t e 5t 5 te 5t 5 e 5t [ 4 5 e e 0 t lim t = lim e 5t t = 0 by LH. 5e 5t t lim t t = lim e 5t t = lim 5e 5t t = 0 by LH. 5e 5t So lim t x e 5x dx = [ 4 5 e e 0 5 e 0 ] is finite. The original series converges. 5 e 0 ] 0

11 Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= only if the improper integral (i). If n= (ii). If n= f (x) is not necessarily positve on [, ) because of the cos function. So Integral test does not apply. Choose D.

12 Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= only if the improper integral (i). If n= (ii). If n= f (x) = x(ln x) p. If p =, then f (x)dx = x(ln x) dx. Let u = ln x, then du = x dx. Then x ln x dx = u du = ln u = ln ln x. x ln x dx = lim t x ln x dx = lim t [ln ln t ln ln ] =. So the original series also diverges. If p, then f (x)dx = x(ln x) p dx. Let u = ln x, then du = x dx. Then x(ln x) p dx = u p du = lim t x(ln x) p dx = lim t p+ (ln x) p ] t = lim t [ p+ (ln t) p p+ (ln) p ]. The only term with t is p+ = u p p+. (ln x) p (ln t) p. lim t (ln t) p = 0 if p > 0 or p > (DNE) if p < 0 or p < So the original series converges if p> and diverges if p.

13 (Remainder Estimate for the Integral Test ). The reminder is R n = s s n = a n+ + a n where s n is the n-th partial sum and s = lim n s n. Suppose f (k) = a k, where f is a continuous, positive, decreasing function for x n and If R n s s n, then s n + n+ f (x)dx s s n + n f (x)dx (a) s 0 = (b) dx = x 5 n+ n 4x 4. n+ f (x)dx R n dx = lim x 5 t dx = lim x 5 t [ + ] = 4t 4 4() t dx = lim x 5 t 0 dx = lim x 5 t [ + ] = 4t 4 4(0) So s s (c) R n n x 5 dx = 4(n) 4 < n > n f (x)dx 3

14 If p = 0, =. n 0 If p > 0, x p dx converges if p> and diverges if p. So n p converges if p> and diverges if 0 < p. 4