The integral test Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f (n). Then the series n= a n is convergent if and only if the improper integral f (x)dx is convergent. Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 43 / 64
Example 2. For what values of p is the series n= n convergent? p Solution: Use the integral test. We know from Chapter 7.8 Indefinite Integral that x dx is convergent if p >, and p divergent if p. Conclusion: The p-series n= n p is convergent if p >, and divergent if p. Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 44 / 64
Example 3. Determine if the series n= diverges. Solution: Use the integral test. n 3 +4 converges or Consider f (x) =. It is positive, continuous and decreasing. x 3 +4 Why is it decreasing? Because x 3 + 4 is increasing. Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 45 / 64
Now x 3 + 4 dx x 3 dx. By Chapter 7.8, the indefinite integral on the right hand side is convergent. Thus dx is also convergent. So the series is x 3 +4 convergent. Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 46 / 64
Example 4. Determine if the series n= ln n n converges or diverges. Solution: Use the integral test. f (x) = ln x x is positive continuous for x >. It is not obvious if f is decreasing or not. Thus we compute f. f (x) = ln x x 2, thus f (x) < 0 when ln x >, that is x > e. Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 47 / 64
So we can apply the integral test (on [e, )): e ln x t dx = lim x t e ln x dx = lim x t We know e = 2.782882846 Thus n=3 ln n n this implies n= ln n n diverges. (ln x) 2 t e =. 2 diverges. Of course, Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 48 / 64
Remark: We learn from this example that to determine the convergence or divergence, we don t mind the first finite number of terms in the series. Thus in the integral test, it is only necessary to assume f (x) is decreasing near infinity. Namely, there exists a constant A, such that f (x) < 0 for x > A. Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 49 / 64
Example 5. Determine if the series n= n2 e n3 converges or diverges. Solution: Use the integral test. Use f (x) = x 2 e x3. f is positive continuous. Is it decreasing? if x. f (x) = e x3 x(2 3x 3 ) < 0 Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 50 / 64
Substitute u = x 3. Then x 2 e x3 dx = 3 Thus the series n= n2 e n3 is convergent. e u du = 3 e u = 3 e <. Chapter : Sequences and Series, Section.3 The integral test and estimates of sums 5 / 64
The comparison test The idea of the comparison test is to compare a given series with a series that is known to be convergent or divergent. Example. Determine if the series 3 n is convergent or + 2 n= divergent. Solution: The series 3 n + 2 reminds us of the series 3 n. n= n= The latter is a geometric series with a = 3, r = 3, and thus it is convergent. Chapter : Sequences and Series, Section.4 The comparison test 52 / 64
The comparison test We can compare each term in these two series: 3 n + 2 < 3 n. Therefore So it is convergent. n= 3 n + 2 < 3 n. n= Chapter : Sequences and Series, Section.4 The comparison test 53 / 64
The comparison test The comparison test Suppose a n and b n are series with positive terms. (a) If b n is convergent and a n b n for all n, then a n is also convergent. (b) If b n is divergent and a n b n for all n, then a n is also divergent. Chapter : Sequences and Series, Section.4 The comparison test 54 / 64
The comparison test We must have some known series b n for the purpose of comparison. Usually, we compare with one of the following two series: (a) Geometric series n= a r n converges for r < and diverges for all other values of r; (b) A p-series n= n p converges for p > and diverges for all other values of p; Chapter : Sequences and Series, Section.4 The comparison test 55 / 64
The comparison test Example 2. Determine if the series divergent. n= n 2 + n + 4 is convergent or Solution: The leading order term as n tends to is n 2. Note the p-series Therefore n= n= n 2 + n + 4 n 2. converges (because p = 2 > ). n2 n 2 + n + 4 converges. Chapter : Sequences and Series, Section.4 The comparison test 56 / 64
The comparison test Example 3. Determine if the series divergent. n= ln n + n is convergent or Chapter : Sequences and Series, Section.4 The comparison test 57 / 64