Lecture 25/26 : Integral Test for p-series and The Comparison test

Transcription

1 Lecture 5/6 : Itegral Test for p-series ad The Compariso test I this sectio, we show how to use the itegral test to decide whether a series of the form (where p =a a ) coverges or diverges by comparig it to a improper itegral. Serioes of this type are called p-series. We will i tur use our kowledge of p-series to determie whether other series coverge or ot by makig comparisos (much like we did with improper itegrals). Itegral Test Suppose f() is a positive decreasig cotiuous fuctio o the iterval [, ) with f() =. The the series = is coverget if ad oly if f()d coverges, that is: If f()d is coverget, the = is coverget. If f()d is diverget, the = is diverget. Note The result is still true if the coditio that f() is decreasig o the iterval [, ) is relaed to the fuctio f() is decreasig o a iterval [M, ) for some umber M. We ca get some idea of the proof from the followig eamples: We kow from our lecture o improper itegrals that d coverges if p > ad diverges if p. p Eample I the picture below, we compare the series = to the improper itegral d. We see that s = + = < + d = + =. Sice the sequece {s } is icreasig (because each > 0) ad bouded, we ca coclude that the sequece of partial sums coverges ad hece the series i= coverges. NOTE We are ot sayig that i= = d here.

2 Eample I the picture below, we compare the series = to the improper itegral d. k= = This time we draw the rectagles so that we get s > s = Thus we see that lim s > lim ad hece sice > d d. However, we kow that d grows without boud d diverges, we ca coclude that k= p-series also diverges. We ca use the result quoted above from our sectio o improper itegrals to prove the followig result o the p-series, i=. p = p coverges for p >, diverges for p. Eample Determie if the followig series coverge or diverge: =, 5, = 5, =0 =00 5,

3 Compariso Test As we did with improper itegral, we ca compare a series (with Positive terms) to a well kow series to determie if it coverges or diverges. We will of course make use of our kowledge of p-series ad geometric series. = p coverges for p >, diverges for p. ar coverges if r <, diverges if r. = Compariso Test Suppose that ad b are series with positive terms. (i) If b is coverget ad b for all, tha is also coverget. (ii) If b is diverget ad b for all, the is diverget. Proof Let s = a i, t = i= b i, i= Proof of (i): Let us assume that b is coverget ad that b for all. Both series have positive terms, hece both sequeces {s } ad {t } are icreasig. Sice we are assumig that = b coverges, we kow that there eists a t with t = = b. We have s t t for all. Hece sice the sequece of partial sums for the series = is icreasig ad bouded above, it coverges ad hece the series = coverges. Proof of (ii): Let us assume that b is diverget ad that b for all. Sice we are assumig that b diverges, we have the sequece of partial sums, {t }, is icreasig ad ubouded. Hece sice we are assumig here that b for each, we have s t for each. Thus the sequece of partial sums {s } is ubouded ad icreasig ad hece diverges.

4 Eample Use the compariso test to determie if the followig series coverge or diverge: = /, = /, = +, =, = l, =! 4

5 Limit Compariso Test Suppose that ad b are series with positive terms. If lim = c b where c is a fiite umber ad c > 0, the either both series coverge or both diverge. a Proof Let m ad M be umbers such that m < c < M. The, because lim b a N for which m < a b < M for all > N. This meas that = c, there is mb < < Mb, whe > N. Now we ca use the compariso test from above to show that If a coverges, the mb also coverges. Hece mb = b m coverges. O the other had, if b coverges, the Mb also coverges ad by compariso coverges. Eample Test the followig series for covergece usig the Limit Compariso test: = = /, = , = ( + ), = + +, = ( π si. ) = e, 5