Math 231E, Lecture 25. Integral Test and Estimating Sums

Transcription

1 Math 23E, Lecture 25. Integral Test and Estimating Sums Integral Test The definition of determining whether the sum n= a n converges is:. Compute the partial sums s n = a k, k= 2. Check that s n is a convergent sequence, i.e. s n = L, where L is finite. However, this is typically quite difficult to do. The challenge tends to be part # of the above. So we state the integral test: Theorem.. Suppose that f(x) is a continuous, positive, decreasing function on [, ). Let a n = f(n). Then converges if and only if converges. In the case that both converge, we have ( ) f(x) f(x) n= a n a n n= ( a + ) f(x). Note that this allows us to check the convergence of a series without having to compute it! 2 p-test for series Example 2.. Does the harmonic series n= n converge? Let us use the integral test. Note that x is a divergent integral from the p-test for integrals, so therefore the harmonic series diverges.

2 Example 2.2. Does the series n= n 2 converge? Let us use the integral test. Note that =, x2 so is a convergent integral from the p-test for integrals, so therefore the series converges. You perhaps already see the pattern, and this gives the p-test for series: The series converges if p > and diverges if p. np n= This is just inherited from the p-test for integrals, and noticing that /x p is continuous, positive, and decreasing on [, ). Note! There is something to be careful about here. Let us consider the sum We know it is convergent since n= n 2. x 2 =. However, what is not true is that the sum and the integral have the same value! For example, we might think that n 2 =, but this is false. In fact, n= n= n 2 = π2 6, which is a fact that is a bit beyond the scope of this class, but you will see in MATH 285/286. In fact, let us compare formulas. We know that if p >, then x p = p, () which means that the sums converge. However, the sums don t have such a simple formula. In fact, it is known that and so on. (Trust me on this!) n= n= n= n= n 2 = π2 6, n 4 = π4 90, n 6 = π6 945, n 8 = π8 9450, 2

3 However, notice that I only put even powers up there. In fact, there is no known formula for any of the odd powers! So, for example, the sum n 3 (2) n= is known to be finite and is a well-studied constant (known as Apéry s constant) but there is no known formula for it. 3 Estimating sums Let s say that we have a convergent series a n = S, n= and we want to approximate its value. If we have a formula for it, great, but usually we do not have a formula for it. The best thing we can do is compute the partial sum s n at some n, and then hopefully it is close to the actual sum. So let us try to estimate the error, or the remainder. If we write R n = S s n, then we would like a formula for this. Theorem 3.. Let f(x) be a continuous, positive function that is decreasing for all x n. Then n+ f(x) R n Example 3.2. Let us say that we want to compute s = n= n f(x) n 3 (3) with and error of no more than 0 4, or How many terms do we need to compute? From the theorem above, we have R n x 3 = 2n 2. So as long as n 2n 2 < 0 4 n 2 > 0 4 n > 00, then our error is less than Of course, we should probably use a computer for this, but it s definitely something a computer can do. In fact, we can compute that 00 n= n 3 =.2020 whereas the actual sum is.20206, so we re only off by

4 4 Proof of Test Proof of Integral Test. Converges. Let us first assume that the integral of f(x) converges. Consider the quantity f(x), the area under the curve from to n. Let us consider a Riemann sum approximation to this area, with x =. If we use left Riemann sums, we obtain f() + f(2) + + f(n ) = a + a a n, and since f is decreasing, left sums are always no smaller than the integral, so we have a + a a n If we now consider right Riemann sums, we obtain f(2) + f(3) + + f(n) = a 2 + a a n. Since f is decreasing, right sums are always no larger than the integral, so we have a 2 + a a n Since the integral of f(x) is convergent, this means that f(x) = f(x). (4) f(x). (5) f(x) exists and is finite. And, since f(x) 0 and using (5), we know that a k k=2 This means that the series of partial sums f(x) s n = k= a k is bounded above (in fact by a + f(x) ). Moreover, notice that s n+ s n = a n+ 0, f(x). so s n is increasing. Since s n is increasing and bounded above, by the Monotone Convergence Theorem the it s n exists and is bounded (again by a + f(x) ). Diverges. If the integral of f(x) diverges, since f(x) 0 this means that f(x) =. 4

5 But from (4) we have which means that so the sum diverges. s n s n f(x), f(x) =, 5