11.6: Ratio and Root Tests Page 1. absolutely convergent, conditionally convergent, or divergent?

Transcription

1 .6: Ratio and Root Tests Page Questions ( 3) n n 3 ( 3) n ( ) n 5 + n ( ) n e n ( ) n+ n2 2 n Example Show that ( ) n n ln n ( n 2 ) n + 2n 2 + converges for all x. Deduce that = 0 for all x. Solutions ( 3) n n 3 We identify a n = ( 3)n n 3. The a n contains a power involving n, so we should try the root test. ( ( a n ) /n ( 3) n ) /n n 3 ( 3 n /n ) n 3 3 n 3/n

2 .6: Ratio and Root Tests Page 2 So we need to know what happens to n 3/n as n. This will turn out to require logarithms to solve. n3/n 0 indeterminate power y = n 3/n ln y = ln n 3/n = 3 n ln n 3 ln y n ln n indeterminate quotient Now we should convert to the reals, since we want to use L Hospital s Rule to evaluate this integral. ln y x = ln x 3 x x indeterminate quotient = /x x using L Hospital s Rule = 0 We want the it y x x eln y = e x ln y = e 0 = So, since we had constructed the real function x 3/x from the discrete n 3/n, we can also say n3/n =. Therefore, we have ( a n ) /n 3 n = 3 3/n = 3 > so the series a n diverges by the root test. You could also use the ratio test to show the series diverges. a n+ a n ( 3) n+ (n + ) 3 n 3 ( 3) n 3n 3 (n + ) 3 ( ) 3 n = 3 n + = 3 ( + /n ) 3 ( ) 3 = 3 = 3 > + 0 The series a n diverges by the ratio test.

3 .6: Ratio and Root Tests Page 3 ( 3) n We identify a n = ( 3)n. The a n contains a factorial, so we should try the ratio test. a n+ a n ( 3) n+ (n + )! ( 3) n 3 (n + ) = 0 < The series a n is absolutely convergent by the ratio test. We identify a n = ( )n 5 + n. ( ) n 5 + n The a n is alternating, so we should try the alternating series test. For the alternating series test, we also need to identify b n = a n = 5 + n. Since b n+ = 5 + n + = 6 + n < 5 + n = b n the first condition for the alternating series test is satisfied. Since b n = 0, the second condition for the alternating series test is satisfied. 5 + n Therefore, by the alternating series test, the series a n converges. But we need to check the convergence of the series b n to determine if the series a n is conditionally convergent (that is, convergent due to the fact that it alternates). Let s use the ratio test to check the series b n. a n+ a n 5 + n 6 + n 5/n + 6/n + = = so the ratio test fails. All this means is we can t use it.

4 .6: Ratio and Root Tests Page 4 Let s try a it comparison test instead. Let s compare to the divergent p-series c n = /n. ( ) c n 5 + n 5 b n n n + = > 0 and finite. Therefore, the since the comparison series c n was divergent, the series b n is also divergent. Therefore, a n is conditionally convergent since a n converges and a n = b n diverges. ( ) n We identify a n = ( )n. ( ) a n+ (n+)! a n ( ) (n + )! n + = 0 < so the series a n is absolutely convergent by the ratio test. We identify a n = e n. e n a n+ a n e (n+) (n + )! e n e (n + ) = e (n + ) = > so the series a n diverges by the ratio test. ( ) n+ n2 2 n We identify a n = ( ) n+ n2 2 n.

5 .6: Ratio and Root Tests Page 5 a n+ a n (n + ) 2 2 n+ (n + )! 2(n + ) 2 (n + )n 2 n + = 2 n 2 ( = 2 n + ) n 2 = 2(0 + 0) = 0 < n 2 2 n so the series a n is absolutely convergent by the ratio test. We identify a n = ( )n n ln n. ( ) n n ln n If we try the ratio test, it will fail. Instead, since a n is alternating, so we should try the alternating series test. For the alternating series test, we also need to identify b n = a n = n ln n. Since b n+ = (n + ) ln(n + ) < n ln n = b n the first condition for the alternating series test is satisfied. Since b n = 0, the second condition for the alternating series test is satisfied. n ln n Therefore, by the alternating series test, the series a n converges. But we need to check the convergence of the series b n to determine if the series a n is conditionally convergent (that is, convergent due to the fact that it alternates). Use the integral test to determine whether the series b n is convergent or divergent. The integral test requires that we work with f(x), where ) f(n) = a n, and on the interval [2, ), f(x) is: ) continuous, 2) positive, 3) decreasing. Here, f(x) =, which is continuous, decreasing and positive on the interval [2, ). x ln x

6 .6: Ratio and Root Tests Page 6 We can therefore apply the integral test to the series n ln n. 2 f(x) dx = 2 x ln x dx t 2 x ln x dx u = ln x when x = 2, u = ln 2 Substitution: du = x dx when x = t, u = ln t ln t ln 2 ln u du u ln t ln 3 = 2 (ln ln t ln ln 2) =, diverges, since ln ln t as t. Since the integral diverges, the series n ln n diverges by the integral test. Therefore, the series n ln n diverges. Therefore, a n is conditionally convergent since a n converges and a n = b n diverges. We identify a n = ( n 2 ) n + 2n 2. + ( n 2 ) n + 2n 2 + The a n contains a power involving n, so we should try the root test. ( ( ( a n ) /n n 2 + 2n 2 + (( n 2 + 2n 2 + n 2 + 2n /n /n 2 = = 2 < ) n ) /n ) n ) /n so the series a n is absolutely convergent by the root test. Example Show that converges for all x. Deduce that = 0 for all x.

7 .6: Ratio and Root Tests Page 7 We identify a n = xn. a n+ a n + (n + )! x n + = x n + = x (0) = 0 < so the series a n is absolutely convergent by the ratio test, for any value of x. Since the series converges, it must be true that the terms in the series are approaching zero (by Theorem.2.6). Therefore, we know that for all values of x, = 0. This is an important result that we will use later. Check out Equation.0.0.