5.3 The Integral Test and Estimates of Sums

Transcription

1 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch 5.3 The Itegral Test ad Estimates of Sums The ext few sectios we lear techiques that help determie if a series coverges. I the last sectio we were able to fid the sum of the series. It s difficult to fid the sum of a series. We ll sped most of our time ow just determiig if the series coverges. Cosider the followig series = = We have o simple formula for the sequece of partial sums S. We re goig to compare the terms of a sequece to the graph of y = x. So the area of each of these rectagles is a term i the series. If we were to add up the area of all these rectagles, we would have the sum for the series. Sice we re oly cocered ow with kowig if it coverges to a fiite umber, we re oly goig to focus o its covergece. Notice that the area of rectagles is less tha the area 4

2 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch therefore, we have the followig relatio = + x dx < + x dx The atural questio at this poit is, does ad we kow how to evaluate those. dx coverge? It s a improper itegral x t dx = lim x x dx = lim t x = lim t + = 0 + = So = < + x dx < + = Sice we kow our series is made up of all positive terms ad we just showed the sum ca t be more tha, we coclude the series coverges. Example 5.3. Let s try to use this idea to determie if Let s compare it to the followig graph of y = x. = coverges. 4

3 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch meas Notice how the area of the each rectagle is larger tha the area uder y = x. This ad = > dx x x dx = lim t = So our series must add to more tha what the itegral sums to ad we just showed that it diverges to. Therefore, our series = x diverges to Why did we choose rectagles above the graph of y = x. Suppose we choose to make our rectagles below y = x, as show below. 43

4 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch This shows that the series sums to a umber less tha = x dx The problem is x dx =. So we really ca t say aythig about the series. Summig to a umber less tha does t mea aythig. The method we used is called the Itegral Test. It ivolves relatig the terms of a series to a improper itegral The Itegral Test Suppose f is cotiuous, positive, decreasig fuctio o [, ) ad let a = f(). The the series a coverges if ad oly if. If f(x) dx coverges. This meas f(x) dx coverges, the a coverges. 44

5 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch. If f(x) dx diverges, the a diverges. Example 5.3. Does the followig series coverge? =. Let s first determie if the sequece a = ( + ) = coverges to 0. ( + ) lim ( + ) lim 4 = 0 Sice the sequece a 0, it s possible the series coverges.. Check to make sure it s decreasig. I order to use the itegral test, the sequece must always be decreasig. We figure this out by lookig at the derivative. Let f(x) = x (x + ). Usi the quotiet rule to fid f (x), we get f (x) = 3x (x + ) For sufficietly large x values, f (x) < 0, meaig the origial fuctio f(x) is always decreasig. 3. The coditios to use the itegral test have bee met. Let s use the itegral test ow. 45

6 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch x t x dx = lim (x + ) (x + ) dx Let u = x +, ad du = x dx t u = lim dx = lim t u = lim t + 4 = = 4 Sice. x dx coverges, the itegral test cocludes the series (x + ) ( + ) coverges 5.3. Covergece of p-series This will be a extremely useful series. I later sectios, we use this series quite a bit. Questio: What values of p is p coverget? It s a pretty simple check. We use the itegral test to determie the covergece. First, we eed to make sure the coditios have bee satisfied to use the itegral test.. Is a = p > 0 for? Yes, a > 0 for all values of p as log as.. Is a decreasig? Let a = f() = p. Sice f () = p p+ < 0 for, we coclude a is always decreasig. 46

7 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch 3. The coditios are met, so let s use the itegral test. For what vaules of p does dx coverge? xp We showed back i the Improper Itegral sectio that x p dx = p, p > 0, p Therfore, the p-series, coverges whe p >. Like I said...simple. p Example Cosider the followig p series. Which oes coverge?.. This series coverges sice p = 6 >. 6. /4. First, rewrite it as a p-series. which diverges because p = 4 <. / =. This series diverges because p = 3 < /. First, rewrite it as a p-series. = / Sice the first p-series has p =.00 > ad the secod series has p = 3/, they both coverge. 47

8 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch Example Determie if the followig series coverges = e. First, let s determie if e is decreasig ad coverges to 0. f () = e e = e ( ) < 0 for all > So the sequece is decreasig. Now we check to see if it coverges to 0. lim e = lim e = lim LH = e = 0. The sequece a = e satisfies the requiremets to use the itegral test. Let s use it ow. We use Itegratio by Parts xe x dx Let u = x, du = dx, dv = e x dx v = e x We ll drop the bouds for ow xe x dx = xe x = xe x + e x dx e x dx = xe x e x 48

9 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch Therefore, xe x dx = lim t xe x dx = lim xe x e x t = lim te t e t ( e e ) = 0 0 ( e ) = e You ca verify by usig L Hospitals Rule. lim te t = 0 3. Now that we showed the itegral coverges, the itegral test cocludes the series e coverges = Example Determie if the followig series coverges l. The sequece a = decreases ad coverges to 0. At this poit, somethig like l this should be easy to idetify.. Sice the sequece satisfies the requiremets for the itegral test, let s use it ow. x l x dx Let u = l x, du = x dx. Usig this substitutio we have l u du = l u l = l(l ) = 49

10 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch 3. Sice the itegral diverges, the Itegral Test cocludes the followig series = l diverges The last part of this sectio is to discuss a way to approximate a series. Suppose you fid the -th partial sum S. Sice this is t the true sum, there is a remaider amout R such that S = a + a + a 3 + a a }{{ + a } + + a + + a }{{} S S is probably a decet approximatio, but sometimes we wat a bit of accuracy. If the sequece a satisfies the itegral test, the R R = a + + a + + a Example Fid the 5-th partial sum of = e We add up the first 5 terms of the sequece a = e. f(x) dx S 5 = e + 4 e + 9 e e + 5 = e5 Suppose I wat a partial sum that s accurate up to This meas R We eed to fid f(x) = x e x = x e x x x e x dx. Usig itegratio by parts, you ca verify that 50

11 5.3 The Itegral Test ad Estimates of Sums Bria E. Veitch Now we just have to solve for x e x dx = e ( + + ) for. e ( + + ) < There really is t a easy way to solve this. I recommed just try pluggig i some values. = 5 e 5 (5 + (5) + ) = > = 0 e 0 (0 + (0) + ) = > = 5 e 5 (5 + (5) + ) = > = 6 e 6 (6 + (5) + ) = > = 7 e 7 (7 + (7) + ) = > = 8 e 8 (8 + (8) + ) = < Therefore, we eed to add up the first 8 terms to be accurate to withi = e = e + 4 e + 9 e = e8 5