Math152 Week 10 11.5

MATH152 WEEK 10

11.5 - Alternating Series We can test convergence of an alternating series by the Alternating Series Test:

∞ X n−1 If the alternating series (−1) bn = b1 − b2 + b3 − b4 + b5 − · · · satisﬁes: n=1 1. bn+1 ≤ bn for all n,

2. lim bn = 0, n→∞

then the series is convergent. (Here bn > 0.) In other words, if the absolute values decrease and go to 0, an alternating series is convergent.

Examples:

(1) Determine whether the following series converge or diverge.

∞ X 1 (a) (−1)n−1 . n n=1

1 Here bn = n , and clearly both conditions 1 and 2 of the Alternating Series Test are satisﬁed. It follows that the series converges. (Note that this test does not tell you what the series converges to.)

∞ X (−1)n 3n (b) . 4n − 1 n=1

3n 3 Here bn = 4n−1 . The limit of the sequence bn as n approaches inﬁnity is 4 , hence condition 2 is not satisﬁed. This means we cannot conclude convergence by the Alternating Series Test. On the other hand, by the Divergence Test, this series diverges.

∞ X (−1)n+1 n2 (c) . n3 + 1 n=3 1 n2 Here bn = n3+1 . One can see that this sequence must eventually be decreasing, since n3 grows faster than n2. Thus it must eventually satisfy condition 1. It also satisﬁes condition 2, for the same reason. Thus the series is convergent by the Alternating Series Test.

P n−1 Suppose an alternating series s = (−1) bn, where bn > 0, satisﬁes the conditions in the Alternating Series Test. Then there is a very simple estimate for the n-th remainder Rn = s − sn, called the Alternating Series Estimation Theorem:

|Rn| ≤ bn+1.

Example:

(2) Consider the series ∞ X (−1)n . n! n=0 Using the Alternating Series Estimation Theorem, ﬁnd the smallest value of n such 1 that |R | ≤ . n 100

1 Here bn = n! . Since n! increases with n, bn is a decreasing sequence and so satisfies condition 1. It also goes to 0 as n → ∞ and so satisfies condition 2. Thus the series converges and we can apply the Alternating Series Estimation Theorem. 1 We want to find the smallest n, given by this theorem, such that |Rn| ≤ 100 . It is 1 suffucient to find n such that bn+1 ≤ 100 , in other words: 1 1 ≤ . (n + 1)! 100 This is equivalent to (n + 1)! ≥ 100. The smallest n for which this occurs is n = 4, so that (n + 1)! = 5! = 120.

2 11.6 - Absolute Convergence and the Ratio Test P P A series an is called absolutely convergent if the series of absolute values |an| is convergent.

Examples: (1) Which of the following series absolutely convergent?

∞ X (−1)n−1 (a) . n n=1

To say that this series is absolutely convergent is to say that the series ∞ n−1 ∞ X (−1) X 1 = n n n=1 n=1 is convergent. But this would be false, since the harmonic series diverges. Thus the original series is not absolutely convergent.

∞ X (−1)n−1 (b) . n2 n=1

To say that this series is absolutely convergent is to say that the series ∞ n−1 ∞ X (−1) X 1 = n2 n2 n=1 n=1 converges. This is true, for it is a p-series with p = 2.

∞ X (−1)n−1 n (c) . n4 + 2 n=3

To say that this series is absolutely convergent is to say that the series ∞ n−1 ∞ X (−1) n X n = n4 + 2 n4 + 2 n=3 n=3 P∞ n converges. This is true, by comparison with the convergent series n=3 n4 .

∞ X (−1)n−1(πn4 − n3 + 1) (d) √ . 10 6 n=3 n − n + 3 3 To say that this series is absolutely convergent is to say that the series

∞ n−1 4 3 ∞ 4 3 X (−1) (πn − n + 1) X |πn − n + 1| √ = √ 10 6 10 6 n=3 n − n + 3 n=3 n − n + 3

converges. We can show that this is false, by applying the Limit Comparison P∞ πn4 Test to this series and the divergent series n=3 n5 . as in last lecture. Therefore the series is not absolutely convergent.

Absolute convergence is a stronger condition than convergence:

X If a series an is absolutely convergent, then it is convergent.

Note that the converse is not in general true. What is an example?

The term conditional convergent means convergent but not absolutely convergent.

Example: ∞ X cos n (3) Determine whether the series is convergent or divergent. n2 n=1

Here the problem actually simpliﬁes if we test it for absolute convergence. To say that this series converges absolutely is to say that

∞ X | cos n| n2 n=1

converges. But since | cos n| ≤ 1 for all n, we have

∞ ∞ X | cos n| X 1 ≤ . n2 n2 n=1 n=1

Since the series on the right converges, the series on the left also converges, by the Comparison Test. It follows that the original series converges absolutely and hence also converges in the usual sense.

4 P∞ One ﬁnal test we will consider is the Ratio Test. Consider a series n=1 an.

an+1 If lim = L < 1, then the series is absolutely convergent. n→∞ an

an+1 If lim = L > 1 or is inﬁnite, then the series is divergent. n→∞ an

an+1 If lim = 1, then the Ratio Test is inconclusive. n→∞ an

Examples:

∞ X n3 (4) Test the series (−1)n for absolute convergence. 3n n=1

(−1)nn3 Let us apply the Ratio Test with an = 3n . We have 3 n 3 an+1 (n + 1) 3 1 n + 1 = n+1 · 3 = . an 3 n 3 n Now divide the top and bottom of the fraction inside the parentheses by the largest term, n, to get 1 1 3 1 + . 3 n 1 This expression approaches 1/3 as n → ∞, since n → 0. Since 1/3 < 1, the given series is absolutely convergent by the ﬁrst statement in the Ratio Test.

∞ X nn (5) Test the series for convergence. n! n=1

In general, when asked to test for convergence, one way we can proceed is to prove absolute convergence, since this would imply convergence. In this case these two types of convergence are actually equivalent, as the terms of the series are already non-negative. n Let us apply the Ratio Test with an = n /n!. We have n+1 n n an+1 (n + 1) n! (n + 1)(n + 1) ) n! n + 1 = · n = · n = . an (n + 1)! n (n + 1)n! n n Again we divide the fraction by the dominant term appearing in the denominator or numerator, which is n. This gives 1 n 1 + , n which in fact approaches e as n → ∞. Thus the series is divergent by the Ratio Test.

5 1 n Remark (not assessable): One way to see that e = limn→∞ 1 + n is as fol- 0 1 lows. Let f(x) = ln x. Then f (x) = x . Using the deﬁnition of the derivative as a limit, we compute f(1 + x) − f(1) ln(1 + x) − ln 1 f 0(1) = lim = lim = lim ln(1 + x)1/x. x→0 x x→0 x x→0 This is the same as 1 x 1 n lim ln 1 + = lim ln 1 + . x→∞ x n→∞ n 0 1 0 1 But since f (x) = x , we have f (1) = 1 = 1. Exponentiating gives the desired result.

P (6) Using the ratio test, what can you say about an in each of the following cases?

an+1 (a) lim = 8. n→∞ an

The ratio test tells us that the series diverges.

an+1 (b) lim = 0.8. n→∞ an

The ratio test tells us that the converges absolutely.

an+1 (c) lim = 1. n→∞ an

The ratio test gives us no information in this case.

∞ X 1 (7) What does the ratio test say about the series ? n2 n=1

Here, the ratio is 2 an+1 n = 2 . an (n + 1) This is a sequence that converges to 1 as n → ∞. Thus the ratio test yields no information in this case (even though we know that the series does converge absolutely because it is a p-series with p = 2).

6 ∞ X 1 (8) What does the ratio test say about the series ? n n=1

Here, the ratio is

an+1 n = . an n + 1 This is a sequence that converges to 1 as n → ∞. Thus the ratio test yields no information in this case (even though we know that the series diverges because it is a p-series with p = 1).

∞ X n (9) Use the Ratio Test to determine whether the series is convergent. 5n n=1

The ratio here is n an+1 (n + 1)5 n + 1 = n+1 = . an n · 5 5n 1 This is a sequence that converges to 5 . Thus by the Ratio Test, the series is absolutely convergent, and hence convergent.

More Exercises:

∞ X (−1)n−1 (10) For what values of K is the series convergent? nK+5 n=1

(Hint: Look at the case K > −5 ﬁrst, using the Alternating Series Test.)

∞ X n6 (11) Test the series (−1)n for convergence. n6 + n3 + 1 n=1

n n6 (Hint: Disregard the (−1) at ﬁrst, and compare to the series n6 .)

∞ X 1 (12) Test the series (−1)n+1 √ for convergence. n=1 n + 5

7 P (13) What can you say about the series an in each of the following cases?

an+1 (a) lim = 100. n→∞ an

an+1 (b) lim = 0.01. n→∞ an

an+1 (c) lim = e. n→∞ an

∞ X 1 (14) What does the ratio test say about the series ? n7 n=1

∞ X (15) What does the ratio test say about the series n? n=1

∞ X n (16) Use the Ratio Test to determine whether the series is convergent. en n=1

8 11.8 - Power Series A power series is a series of the form ∞ X n 2 3 cnx = c0 + c1x + c2x + c3x + ··· n=0

Here x is a variable and the cn are called coeﬃcients. Notice that such a series may not converge for all values of x. Another way to view a power series is as a polynomial with inﬁnitely many terms:

2 n f(x) = c0 + c1x + c2x + ··· + cnx + ···

Examples:

∞ X (1) xn = 1 + x + x2 + ··· + xn + ··· . n=0 This power series is a geometric series with a = 1 and r = x.

∞ n X 1 1 1 1 1 (2) = 1 + + + + + ··· 2 2 4 8 16 n=0 1 This is the above power series evaluated at x = 2 .

∞ X n 2 (3) cn(x − a) = c0 + c1(x − a) + c2(x − a) + ··· n=0 This is called a power series centered at a. It is a power series in the variable (x − a) and is not necessarily a geometric series (why?).

For what values of x does the ﬁrst power series converge?

∞ X (4) For what values of x is the series n!xn convergent? n=0 It is convenient to use the Ratio Test to determine convergence of power series. n With an = n!x , we have n+1 an+1 (n + 1)!x lim = lim n = lim (n + 1)|x| = ∞. n→∞ an n→∞ n!x n→∞ Thus by the Ratio Test, the series is divergent for any x 6= 0. When x = 0, we see that the series converges, since there is only one non-zero term.

9 ∞ X (x − 3)n (5) For what values of x is the series convergent? n n=1

(x−3)n Again, let us apply the Ratio Test, with an = n . We have n+1 an+1 (x − 3) n 1 lim = lim · = |x − 3|. n→∞ n→∞ n 1 an n + 1 (x − 3) 1 + n For any ﬁxed value of x, this sequence approaches |x − 3| as n → ∞. By the Ratio Test, if |x − 3| < 1, then the series is absolutely convergent (and hence convergent). This occurs if and only if 2 < x < 4. Again by the Ratio Test, the series diverges if |x − 3| > 1. This occurs if and only if x < 2 or x > 4. When |x − 3| = 1, the Ratio Test yields no information. This occurs when x = 2 or 4. We need to test these two cases separately. When x = 2, the series becomes P∞ (−1)n n=1 n , which is convergent by the Alternating Series Test. When x = 4, the P∞ 1 series becomes n=1 n , which diverges. Thus we conclude that the given series converges if and only if 2 ≤ x < 4.

(6) Find the largest domain of the function of order 0 deﬁned by ∞ X (−1)nx2n J (x) = , 0 22n(n!)2 n=0 called the Bessell function of order 0.

This question is the same question as: find the values of x for which the series in the definition of J0(x) converges. We apply the Ratio Test with (−1)nx2n a = . n 22n(n!)2 We have n+1 2(n+1) 2n 2 an+1 (−1) x 2 (n!) lim = 2(n+1) 2 · n 2n . n→∞ an 2 ((n + 1)!) (−1) x This simplifies to x2 , 4(n + 102) which goes to 0 for any value of x. Thus, by the Ratio Test, the given series converges for all values of x. Thus the largest domain of J0 is R.

10 Suppose we have a power series centered at a, ∞ X n 2 cn(x − a) = c0 + c1(x − a) + c2(x − a) + ··· n=0 It turns out that there are only three possibilities for the convergence of any such series:

1. The series converges only when x = a. 2. The series converges for all x. 3. There is R > 0 such that the series converges if |x − a| < R and diverges if |x − a| > R.

The number R is called the radius of convergence. You can think of the ﬁrst two possibilities as the cases R = 0 and R = ∞ respectively.

Examples:

(7) Find the radius of convergence of the following series.

∞ X (a) n!xn n=0

We showed above that this series converges if and only if x = 0. Thus its radius of convergence is R = 0.

∞ X (b) xn n=0

We know that this is a geometric series that converges if and only if |x| < 1, or −1 < x < 1. The length of this interval is 2, so its radius of convergence is R = 1.

∞ X (x − 3)n (c) n n=1

We showed above that this series converges if and only if 2 ≤ x < 4. The length of this interval is 2, so its radius of convergence is R = 1.

11 ∞ X (−1)nx2n (d) J (x) = 0 22n(n!)2 n=0

We showed above that this series converges for all values of x. Thus its radius of convergence is R = ∞.

Thus a power series centered at a always has some interval of convergence.

(8) Find the interval of convergence of the following series.

∞ X (a) n!xn n=0

The interval of convergence of this series is [0, 0].

∞ X (b) xn n=0

The interval of convergence of this series is (−1, 1). Alternatively you can write this as −1 < x < 1.

∞ X (x − 3)n (c) n n=1

The interval of convergence of this series is [2, 4). Alternatively you can write this as 2 ≤ x < 4.

∞ X (−1)nx2n (d) J (x) = 0 22n(n!)2 n=0

The interval of convergence of this series is (−∞, ∞) = R. 12 More Exercises:

∞ X (−3)nxn (9) Find the radius of convergence and interval of convergence of the series √ . n=0 n + 1

(Hint: Use the Ratio Test, and remember to treat any points where the Ratio Test is inconclusive separately.)

∞ X n(x + 2)n (10) Find the radius of convergence and interval of convergence of the series . 3n+1 n=0

(Hint: Use the Ratio Test, and remember to treat any points where the Ratio Test is inconclusive separately.)