THE INTEGRAL TEST AND ESTIMATES OF SUMS

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1 THE INTEGRAL TEST AND ESTIMATES OF SUMS. Itroductio Determiig the exact sum of a series is i geeral ot a easy task. I the case of the geometric series ad the telescoig series it was ossible to fid a simle formula for the th artial sum s. For most series it will ot be easy to determie such a formula. Eve if we caot determie the exact value, aroximatio will be helful. For this reaso we will examie several tests to determie whether a series is coverget or diverget without fidig its sum. 2. The Itegral Test The first aroach exloits the coectio betwee sequeces, series, ad cotiuous fuctios o the real lie. Cosider the series whose terms are recirocals of the squares of the ositive itegers: 2 = i= There is o simle formula for the sum s of the first terms, but by cosiderig fiite sums, it aears that the series ears.64 as, ad thus it would be coverget. a If we cosider the fuctio y = /x 2, the this series corresods to a very coarse Riema sum with each rectagle is of legth, ad the height is equal to the value of the fuctio y = /x 2 at the right edoit of each iterval. If we exclude the first rectagle, the total area of the remaiig rectagles is smaller tha the area uder the curve y = /x 2 for x which equals (/x2 )dx This itegral is coverget (it has value ), ad it imlies that all artial sums are less tha 2 + dx = 2, x2 ad the artial sums are bouded. Sice all terms are ositive, the artial sums are icreasig, thus the artial sums coverge by the Mootoic Sequece theorem

2 2 THE INTEGRAL TEST AND ESTIMATES OF SUMS ad the series must be coverget. The sum of the series will be less tha 2 so that it ever reaches its uer-boud: 2 = < i= As aother examle, cosider = i= The table of artial sums suggests that these sums are ot aroachig a fiite umber, ad so this might be a diverget series. a To rove this we cosider the fuctio / x ad emloy rectagles of width with height determied by the left edoit, so that the rectagle tos lie above the curve. The total area is greater tha the area of the curve y = / x for x which equals (/ x)dx. As this imroer itegral is diverget, the area uder the curve is ifiite ad so the sum of the series must be ifiite, ad hece it is diverget. This aroach ca be geeralized to give the followig theorem: Theorem 2.. Suose f is a cotiuous, ositive, decreasig fuctio o [, ) ad let a = f(). The the series = a is coverget if ad oly if the imroer itegral f(x)dx is coverget. I.e., If f(x)dx is coverget the = a is coverget. If f(x)dx is diverget the = a is diverget. Remark 2.2. It is ot ecessary that f is always decreasig, oly that f is ultimately decreasig for all x larger tha some value N. Examle 2.3. Test the series = 2 + to determie covergece. Proof. The fuctio x 2 + is cotiuous, ositive, ad decreasig o [, ), ad so by the Itegral test t x 2 dx = lim + t x 2 dx = lim + t arcta(x)]t ( = lim arcta t π ) = π 4 2. As the itegral is coverget, the series test. = x 2 + Examle 2.4. For what values of is the series = is coverget by the Itegral coverget?

3 THE INTEGRAL TEST AND ESTIMATES OF SUMS 3 Proof. If < 0 the lim =. If = 0 the lim =. As lim 0 both of these cases are diverge by the test for divergece. If > 0 the the fuctio is cotiuous, ositive, ad decreasig o [, ). x Better yet, we kow that x dx coverges if > ad diverges if. Therefore, the Itegral test imlies the series coverges whe > ad diverges if 0. This is a very useful result, ad so it will get its ow theorem: Theorem 2.5. The -series. Examle 2.6. Determie if = = +4 2 is coverget if > ad diverget if is coverget. Proof. As f = x+4 x is cotiuous ad ositive for x, if it is decreasig we may 2 aly the Itegral test; takig the derivative f (x) = x2 2 x = 3x + 6 x 2x 7 2 we see that this is the case for all x. This itegral is coverget sice x + 4 t ( x + 4 x 2 dx = lim t x 2 dx = lim + 4 ) t x 2 dx = lim t = 6 [ 2 x 4 x ] t < 0 = lim t x 3 2 ( 2 t 4r ), ad so the series is coverget as well. Alteratively, we could have oticed that for very large the series will be equivalet to which coverges by the -series with = >. 3. Estimate the Sum of a Series Now that we ca show a series is coverget, if a give series has a fiite sum, s, the ext questio is to ask what is its aroximate value. We could use a artial sum s a a aroximatio to s as lim s s. To see if our estimate s is accurate eough, cosider the remaider: R = s s = a + + a +2 + a +3 + The remaider R is the error made whe s is used to aroximate the total sum. We will determie bouds o R usig a similar idea emloyed i the Itegral test. Assumig that f is decreasig o [, ), we ca comare the areas of the rectagles with the area uder y = f(x) for x > this imlies ad R = a + + a +2 + R = a + + a This roves the followig error estimate + f(x)dx f(x)dx.

4 4 THE INTEGRAL TEST AND ESTIMATES OF SUMS Theorem 3.. Remaider Estimate for the Itegral Test. Suose f(k) = a k, where f is a cotiuous, ositive, decreasig fuctio for x ad a is coverget. If R = s s the + f(x)dx R f(x)dx Examle 3.2. Aroximatig the sum of the series usig the first te terms 3 3 s 0 = = The size of the error for this value at most How may terms are required to esure that the sum is accurate to withi ? Proof. Accuracy to withi meas that we have to fid a value of such that R As we wat the iequality Solvig this we fid R x 3 dx = < > 0.00 = 000 or > Thus, 32 terms are eeded to esure this accuracy. If we wish to boud the actual value of the series, add s N to each side of the iequality for R () s + f(x)dx s s + f(x)dx + sice s + R = s Examle 3.3. Estimate the sum of the series = with = 0 3 Proof. From the last examle we kow that Ad so Usig s we fid x 3 dx = 2 2 s 0 + 2() 2 s s0 + 2(0) s Takig the midoit of this iterval as a aroximatio of s, the the error is at most half the legth of the iterval, ad so.202, with error < =

5 THE INTEGRAL TEST AND ESTIMATES OF SUMS 5 Remark 3.4. Comarig the last two examles, the estimate i () is much better tha the estimate s s. To make the error smaller tha we would have to add the first 32 terms for s, or 0 terms usig ().