Section 11.3: The Integral Test


 Nelson Johnston
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1 Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult tha what we have see. Covergece ca be broke up ito two problems ) Does a series coverge? ad 2) What does it coverge to? To tackle the first problem, we shall develop a whole ew toolbox based o results we have already used i Calculus. The secod problem is much more difficult ad requires more advaced mathematics tha we curretly have. Therefore, our focus for most of this chapter will be o the first problem (though we shall occasioally use the secod).. Usig Itegrals to Determie Covergece  a Example We start with a example. Example.. Show that the sum = Cosider the fuctio f(x) = /x 2 which coicides with this sum for iteger values. We ca approximate the area uder this curve usig a Riema sum. If we take the right had Riema sum with subdivisios of legth, the we get the followig ifiite sum = which is the same as the series we are tryig to fid. If we igore the first term of the sum ad just cosider the ifiite sum =2 the the value of this sum will be strictly bouded by the area uder the graph from to (sice the right had sum produces a uderestimate). But this area ca be estimated usig a improper itegral. Summarizig, we have = = + 2 = + 2 x 2dx.
2 2 I particular, the sum will oly coverge if the itegral Calculatig, we have x 2dx = x =. Therefore, sice the iterval coverges ad is larger tha the sum, the sum must coverge. Hece the sum = = WARNING: Just because it coverges does ot mea we have foud the actual value of the sum. I fact, all we have show is that the sum = + 2 < + 2 x2dx = 2. =2 2. The Itegral Test The method we used i our example approximatig the sum with a itegral ca be applied to lots of other cases to show that a series coverges (observe that we did ot fid what the sum was (it is i fact π 2 /6)). There are just a couple of thigs which are ecessary for it to work: (i) The sequece i the series must be positive for all values past a certai poit. (ii) The sequece i the series must be decreasig after a certai poit (to guaratee that we will be gettig a uderestimate). (iii) The fuctio correspodig to the sequece is cotiuous (so we ca take the Riema sum). If these thigs are satisfied, the we ca apply the followig: Result 2.. (The Itegral Test) Let a be a series with positive terms ad let f(x) be the fuctio that results whe is replaced by x. If f is decreasig ad cotiuous o [a, ) the a ad either both coverge or diverge. a f(x)
3 Observe that we oly start at a  this was illustrated i the example because if we tried to itegrate from 0 to, the itegral will have diverged. Of course, the key poit is that the first few terms will ot affect divergece or covergece  it is the ultimate behavior which couts ad this is measured by the itegral. We illustrate the power of the itegral test with a few examples. Example 2.2. Show that the harmoic series diverges. Observe that the harmoic series agrees with the fuctio f(x) = /x. However, dx = l(x) x which diverges. Sice the itegral diverges, the correspodig series must diverge. Example 2.3. Show that the sum 3 + First observe that sice. Therefore Usig the itegral test, we have so x 3dx = 2x 2 3 = 2, 3 coverges, ad hece so does 3 +. Example 2.4. We defie a p series to be the series p where p is some iteger. Fid all values of p for which this series Clearly if p 0, the this series diverges (because we shall be addig ifiitely may umbers greater or equal to together. Also, the fuctios will ot be decreasig for p < 0, so we could ot apply the itegral
4 4 test ayway! Also, the case p = was cosidered above (it is the harmoic series). Therefore, we may assume that p > 0 ad p. The correspodig itegral will be x pdx = x p x p dx = lim k p k k p = lim k p p The oly part of this limit which could diverge is the term k p, ad this will diverge if ad oly if p > 0 or p <. Therefore, a pseries p coverges if ad oly if p >. Example 2.5. Determie which values of p the series e p If p 0, this clearly diverges, so we assume p < 0. This is a decreasig ad cotiuous fuctio, so we apply the itegral test: e xp dx = lim p k e xp k = lim e kp k p + e p p = e p p This itegral coverges for all p > 0, so the series coverges for all p > Estimatig the Sum of a Series Though out mai cocer is determiig whether a series coverges, we shall briefly discuss how we ca approximate the value of a series which does coverge usig the itegral test. If a series coverges, the the more terms we add up, the closer the value will be to the actual value of the sum. Suppose that the actual value of a series is S ad the th partial sum is deoted by S. The we defie the remaider R = S S, the differece betwee the actual value of the series ad the partial sum. Obviously, the smaller the remaider is, the closer the partial sum is to the actual value of the series. We ca use itegrals to determie the remaider: Result 3.. Suppose f() = a where f is cotiuous, positive, decreasig for x ad a is coverget. If R = S S, the + f(x)dx R f(x)dx.
5 This result basically tells us that the remaider ca be approximated by these two itegrals. We illustrate. Example 3.2. Fid the value of which guaratees that the remaider will be withi 0. of the actual value for the sum. 5 Observe that x 2dx = x = If we choose = 0, the we have R 0 0 so oly 0 terms i the sum are eeded to get the value of withi 0. of the actual aswer. Usig the TI89 sum optio, we have R 0 = Observe that π 2 /6 =.64493, so it is withi 0. of the actual aswer!
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