4¨‰(‰{|}~€‚ƒ„y wzabBň” GkBhcThe Integral TestkBhc+$Introduction to integral comparisonskBhcqjThe method for computing the value of a series is based on taking the limit of a sequence of partial sums.kBhcssbuSkBhcnkBhc  = muS(kBhcnkBhc k = 0kBhc  asbuSkBhck muS(kBhc°kBhc k = 0kBhc  asbuSkBhckkBhc  = tmiLkBhc nĎ°kBhc  SsbuSkBhcnkBhc˙The major drawback to this method is that it is usually very difficult to derive a formula for the partial sum. To work around this problem, we are going to see a number of more indirect methods for determining whether or not series converge and estimatinkBhc?8g their values in those cases in which they do converge. kBhcC etnIFkBhc 1/xkBhcxkBhc°kBhc1/kBhc92When we do the resulting improper integral, we get0tmiLkBhc AĎ°etnIFkBhc 1/xkBhcxkBhcAkBhc1kBhc  = tmiLkBhc AĎ°GrapUkBhc lnVsbAkBhcAkBhc  - lnVsbAkBhc1kBhc  = +°1kBhc?8We conclude from this that the harmonic series diverges.23kBhc|uHere is another example. This time around, lets see if we can guess what might happen before we get into the details.4 muS(kBhc°kBhc k = 1carF<kBhc1kBhckopxEkBhc25kBhcš˛The first thing you should do should be to write a simple integral that looks like the series without thinking too much about the picky details such as the limits of integration.6etnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc1kBhc  = 17kBhc†Given that we now suspect that the series may converge, we ought to try to get a set of rectangles that sit under the graph of kBhcy = 1/xopxEkBhc2kBhc;4. The rectangles depicted below show how to do this.8tciP"ţţ"˙ ˙˙˙˙ĄňMSWD €€˙˙" 0ü ˜ ˜ +8, Geneva (+0 ™ ˜). ™ ˜)5 ™ ™ €€˙˙" Rü ˜ ˜ PW)1 ™ ™ €€˙˙" tü ˜ ˜ o{)1 ™ ˜). ™ ˜)5 ™ ™ €€˙˙" •ü ˜ ˜ “š)2 ™ ™ €€˙˙" ˇü ˜ ˜ ˛ż)2 ™ ˜). ™ ˜)5 ™ ™ €€˙˙" Ůü ˜ ˜ ×Ţ)3 ™ ™ €€˙˙" űü ˜ ˜ ö)3 ™ ˜). ™ ˜)5 ™ ™ €€˙˙" ü ˜ ˜ ")4 ™ ™ €€˙˙"đ ˜ ˜ íő (ô0 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"Ó ˜ ˜ ĐŘ (×1 ™ ™ €€˙˙"ś ˜ ˜ łť (ş1 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"™ ˜ ˜ –ž (2 ™ ™ €€˙˙"} ˜ ˜ z‚ (2 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"` ˜ ˜ ]e (d3 ™ ™ €€˙˙"C ˜ ˜ @H (G3 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"& ˜ ˜ #+ (*4 ™ ™ €€˙˙    & "-*#:###### ### ### "ů~ # # # # # # # # # # # # # Ąś ĚĚĚĚĚĚ1Ő Q0ÔRĄś ĄśĄś ĚĚĚĚĚĚ1R •0˙Q–Ąś ĄśĄś ĚĚĚĚĚĚ1– Ů0•ÚĄś Ąś˙9kBhcőîThis time around the rectangles sit under the curve. We did this by giving each rectangle a height given by the value of the function at the right hand end-point of the interval it sits over. The picture suggests the following comparison.: muS(kBhc°kBhc k = 1carF<kBhc1kBhckopxEkBhc2kBhc  < etnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc0kBhc  = +°;kBhc˙The comparison is not terribly useful, because the obvious integral to compare the series to diverges. This is easy to fix. All we have to do is to split off the first term of the series and use the integral comparison to control the other terms. That meakBhc.'ns using the portion of the curve over kBhc [1,°)kBhc instead of kBhc (0,°)kBhc for our comparison.< muS(kBhc°kBhc k = 1carF<kBhc1kBhckopxEkBhc2kBhc < 1 + etnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc1kBhc = 1 + 1 = 2=kBhc-&Thus we see that the series converges.>kBhcIBUsing an integral comparison to estimate the remainder in a series?kBhc#Suppose we wanted to compute@ muS(kBhc°kBhc k = 1carF<kBhc1kBhckopxEkBhc2AkBhcLEWe could estimate the value of this series by computing partial sums.BkBhcssbuSkBhcnkBhc  = muS(kBhcnkBhc k = 1carF<kBhc1kBhckopxEkBhc2kBhc CkBhc?8Here is a table of partial sums for various values of N.DlbaT\kBhcnkBhc2kBhc4kBhc8kBhc 16kBhc 32kBhc 64kBhcssbuSkBhcnkBhc 1.25kBhc1.4236111111111112kBhc1.527422052154195kBhc1.5843465334449871kBhc1.6141672628279242kBhc1.6294305014088875EkBhc˙This gives us some sort of sense for what the sum will eventually be, but it does not tell us how many terms we need to add up in order to get the answer correct to within a certain predetermined accuracy. To understand how many terms we will need to add kBhc\Uup in order to get that accuracy, it helps to write the series as a sum of two parts.F muS(kBhc°kBhc k = 1carF<kBhc1kBhckopxEkBhc2kBhc  = muS(kBhcnkBhc k = 1carF<kBhc1kBhckopxEkBhc2kBhc  + muS*kBhc°kBhck = n+1carF<kBhc1kBhckopxEkBhc2GkBhc2+actual answer = part we can compute + errorHkBhc‰‚How do we get a sense for how big the error term is? The answer is that we can use some sort of integral comparison to control it.IetnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc?kBhc  < muS*kBhc°kBhck = n+1carF<kBhc1kBhckopxEkBhc2kBhc  < etnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc?JkBhc˙We want to control the error term both from below and from above by means of some integral comparisons. It is clear what form those integrals have to take, so the only detail that remains to be determined is what bounds we need to set for the integrals tokBhcˆ guarantee the comparisons we need. To see what the bounds should be, let us get a little more concrete by assuming a definite value of kBhcnkBhc.KkBhc n = 4LkBhc_XThe first picture below shows how to set up a collection of rectangles to guarantee thatM muS(kBhc°kBhc k = 5carF<kBhc1kBhckopxEkBhc2kBhc  < etnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc4NtciP˘&ö’’ö&˙ ˙˙˙˙ĄňMSWD €€˙˙"ę4ü ˜ ˜ î/ö<, Geneva +/ő4 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"ęVü ˜ ˜ îTö[)5 ™ ™ €€˙˙"ęxü ˜ ˜ îsö)5 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"ę™ü ˜ ˜ î—öž)6 ™ ™ €€˙˙"ęťü ˜ ˜ îśöĂ)6 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"ęÝü ˜ ˜ îŰöâ)7 ™ ™ €€˙˙"ę˙ü ˜ ˜ îúö)7 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"ę!ü ˜ ˜ îö&)8 ™ ™ €€˙˙"ź ˜ ˜ šÁ(Ŕ0 ™ ˜). ™ ˜)0 ™ ˜)5 ™ ™ €€˙˙"Ž ˜ ˜ ‹“(’0 ™ ˜). ™ ˜)1 ™ ™ €€˙˙"` ˜ ˜ ]e(d0 ™ ˜). ™ ˜)1 ™ ˜)5 ™ ™ €€˙˙"2 ˜ ˜ /7(60 ™ ˜). ™ ˜)2 ™ ™ €€˙˙" ˜ ˜ (0 ™ ˜). ™ ˜)2 ™ ˜)5 ™ ™ €€˙˙ ęę! ę"ą # # # # # # # # # "΂ # # # # # # # # # # # # #  ŒĄś ĚĚĚĚĚĚ1ÇéV0ĆęWĄś ĄśĄś ĚĚĚĚĚĚ1ŇWé™0ŃVꚡś ĄśĄś ĚĚĚĚĚĚ1ښéÝ0ٙęŢĄś Ąś ˙OkBhcž—Shifting those rectangles to the right by one unit reverses the inequality and also slightly changes the limits of integration used for the comparison.PetnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc5kBhc  < muS(kBhc°kBhc k = 5carF<kBhc1kBhckopxEkBhc2QtciP˘&ö’’ö&˙ ˙˙˙˙ĄňMSWD ŒĄś ĚĚĚĚĚĚ €€˙˙1ÇWéš0ĆVꛡś ĄśĄś ĚĚĚĚĚĚ1қéÝ0њęŢĄś ĄśĄś ĚĚĚĚĚĚ1ÚŢé!0ŮÝę"Ąś Ąś "ę4ü ˜ ˜ î/ö<, Geneva +/ő4 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"ęVü ˜ ˜ îTö[)5 ™ ™ €€˙˙"ęxü ˜ ˜ îsö)5 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"ę™ü ˜ ˜ î—öž)6 ™ ™ €€˙˙"ęťü ˜ ˜ îśöĂ)6 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"ęÝü ˜ ˜ îŰöâ)7 ™ ™ €€˙˙"ę˙ü ˜ ˜ îúö)7 ™ ˜). ™ ˜)5 ™ ™ €€˙˙"ę!ü ˜ ˜ îö&)8 ™ ™ €€˙˙"ź ˜ ˜ šÁ(Ŕ0 ™ ˜). ™ ˜)0 ™ ˜)5 ™ ™ €€˙˙"Ž ˜ ˜ ‹“(’0 ™ ˜). ™ ˜)1 ™ ™ €€˙˙"` ˜ ˜ ]e(d0 ™ ˜). ™ ˜)1 ™ ˜)5 ™ ™ €€˙˙"2 ˜ ˜ /7(60 ™ ˜). ™ ˜)2 ™ ™ €€˙˙" ˜ ˜ (0 ™ ˜). ™ ˜)2 ™ ˜)5 ™ ™ €€˙˙ ęę! ę"ą # # # # # # # # # "΂ # # # # # # # # # # # # # ˙RkBhc(!Putting this all together we haveSetnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc5kBhc  < muS(kBhc°kBhc k = 5carF<kBhc1kBhckopxEkBhc2kBhc  < etnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc4TkBhc%Computing the integrals we getUcarF$kBhc1kBhc5kBhc  < muS(kBhc°kBhc k = 5carF<kBhc1kBhckopxEkBhc2kBhc  < carF$kBhc1kBhc4VkBhc#Thus, for the special case nkBhc  = 4kBhc we have thatW muS(kBhc°kBhc k = 1carF<kBhc1kBhckopxEkBhc2kBhc  = muS(kBhc4kBhc k = 1carF<kBhc1kBhckopxEkBhc2kBhc  + muS(kBhc°kBhc k = 5carF<kBhc1kBhckopxEkBhc2XkBhc‰‚and the error term is no smaller than 1/5 and no larger than 1/4. Once we have seen how to do all of this for a specific value of kBhcnkBhc=6, the comparisons are easy to generalize to arbitrary kBhcnkBhc.Y carF&kBhc1kBhc n+1kBhc  = etnIzcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhc n+1kBhc  < muS*kBhc°kBhck = n+1carF<kBhc1kBhckopxEkBhc2kBhc  < etnIxcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc°kBhcnkBhc  = carF$kBhc1kBhcnZkBhcFor example, with kBhc n = 64kBhc$ we have from the table above[kBhcSsbuSkBhc 64kBhc = 1.6294305014088875\kBhc&and the estimate is accurate to]carF%kBhc1kBhc 65kBhc < Error with 64 terms