Ifiite Series Defiitio. A ifiite series is a expressio of the form uk = u + u + u + + u + () 2 3 k Where the umbers u k are called the terms of the series. Such a expressio is meat to be the result of addig together ifiitely may umbers. As with improper itegrals, the above expressio ca sometimes be give a meaig ad sometimes ot.
Sice it is physically impossible to add together a ifiite umber of terms, we must attempt to give expressio () meaig through the idea of limit. Ifiite series are used i ordiary arithmetic i the form of decimal expasios. For example whe we write 2/3 = 0.6666666 we are expressig the idea that the ifiite series 0.666 = 6 + 6 + 6 + + 6 0 00 000 0 k + is somehow meaigful ad has the value 2/3.
How might we give meaig to a series? Oe way is to begi by addig together all the terms of the series from term to term. The result is s = u + u + u + + u = u 2 3 k ad is called the th partial sum of the series u k We the focus attetio o the sequece of partial sums s =,,,, s, = = u, u+ u2, u+ u2+ u3,, u+ u2+ + u, { } ss s 2 3
As gets ever larger, the partial sum s represets the sum of more ad more terms of the series. Thus it is reasoable to thik that if the sequece s teds to a limit u (ad therefore gets ever closer to u) the limit u is what we should mea by the sum of all the terms. Thus we are led to the followig defiitio. Defiitio. Suppose that is a series ad that { s } = is the correspodig sequece of partial sums. If the sequece { s } coverges to a limit s, we say that the series = coverges to s ad that s is the sum of the series. If the sequece of partial sums diverges, we say the series diverges, ad therefore has o sum. u k u k
Example. Determie whether the series + + coverges or diverges, ad if it coverges, fid its sum. Solutio. We see that s =, s 2 = 0, s 3 =, s 4 = 0, Sice this sequece oscillates betwee 0 ad, it will ot evetually get ad stay close to ay umber, ad so it has o limit. Thus the series sum. + + does ot coverge ad so has o
Without a formal defiitio, this sequece could, ad did, cause cotroversy. O oe had you ca reaso that the sum should be ( ) ( ) + + = 0 O the other had we could write the series as ( ) ( ) + + + + + = The defiitio shows us that o meaig at all ca be attached to the series.
Geometric Series Oe special type of series occurs frequetly ad is easily hadled. That is the series 2 3 k a+ ar+ ar + ar + ark + = ar 0 This is called a geometric series, ad r is called the ratio. Each term is multiplied by r to produce the ext term. Note here that we have begu the series with a subscript 0 rather tha, ad so started with the 0 th term istead of the first term. Actually we ca begi the series with ay subscript, just as was the case with a sequece, but 0 ad are the most commo startig poits..
The geometric series is hadled by a simple algebraic trick. k The th partial sum of the series ar 0 s = a+ ar+ ar + ar + ar 2 3 the rs = ar+ ar2+ ar3 + ar+ ar + is ad therefore s rs = [ a+ ar+ ar + ar ] ar+ ar + ar + ar + ar We have = a ar+ = a( r+ ). s 2 2 3 + + r = a r if r (it is easily see that the series diverges if r = or r = )
If r >, s does ot coverge. If r <, s does coverge sice powers of r ted to 0. We therefore have If r, the series k ar 0 diverges. If r <, the series coverges ad + r s= lim s = a lim a =. r r Therefore s is the sum of the series.
Example. Determie whether the series 2 k+ 2 3 4 5 2 2 2 8 6 32 = + + + = + + + 3 3 3 3 27 8 243 coverges or diverges, ad if it coverges determie the sum. Solutio. This is a geometric series, sice each term is multiplied by 2/3 to get the ext. We have a = 8/27 ad r = 2/3. Sice the ratio has absolute value less tha, the series coverges to 8 a 27 8 r = 2 = 9 3
Example. Determie whether the series 3 k+ 2 3 4 3 3 3 9 27 8 = 2 2 + 2 + 2 + = + 4 8 6 coverges or diverges, ad if it coverges determie the sum. Solutio. This is a geometric series, sice each term is multiplied by 2/3 to get the ext. We have a = 9/4 ad r = 3/2. Sice the ratio has absolute value greater tha, the series diverges.
Example. Determie whether the series e k 4 5 6 e e e = π π + π + π + 5 coverges or diverges, ad if it coverges determie the sum. Solutio. This is a geometric series, sice each term is multiplied by e/π to get the ext. We have a = e 4 / π 4 ad r = e/π. Sice the ratio has absolute value less tha, the series coverges to 4 4 4 a = e = e π = e. r 4 e 4( π e) 3 π π π ( π e) π
Example. Determie whether the series 3k k 625 5 7 7 = 7 coverges or diverges, ad if it coverges determie the sum. k Solutio. This is a geometric series, sice each term is multiplied by the ratio 625/7 to get the ext. Sice the ratio has absolute value greater tha, the series diverges.
Every decimal that evetually repeats a strig of digits edlessly is said to be repeatig. Every ratioal umber ca be expressed as a repeatig decimal, ad coversely every repeatig decimal coverges to a ratioal umber. Examples of repeatig decimals are..333333333 2.472345345345345 0.5555555 2.4566666666666.00353535 0.9999999
Problem. Fid the ratioal umber that is the sum of the repeatig decimal.333333333 Solutio. This decimal is the same as the ifiite series 3 3 3 3 + + + =+ 0 00 000 0 0 0 The series that follows the is geometric with ratio /0. Thus it coverges to the umber 3 0 = 3 0 The decimal therefore represets 4/3. k
Problem. Fid the ratioal umber that is the sum of the repeatig decimal.99999 Solutio. This decimal is the same as the ifiite series 9 9 9 9 + + = 0 00 000 0 0 0 The series is geometric with ratio /0. Thus it coverges to the umber 9 0 = 0 The decimal therefore represets. k
Problem. Fid the ratioal umber that is the sum of the repeatig decimal 0.595959 Solutio. This decimal is the same as the ifiite series 59 59 59 59 + + = 3 6 9 0 3 0 3 0 0 0 0 The series is geometric with ratio /000. Thus it coverges to the umber 59 000 = 59 999 000 The decimal therefore represets 59/999. k
Problem. Fid the ratioal umber that is the sum of the repeatig decimal 0.222 Solutio. This decimal is the same as the ifiite series 2 2 2 2 + + = 2 4 6 0 2 0 2 0 0 0 0 The series is geometric with ratio /000. Thus it coverges to the umber 2 00 = 2 99 00 The decimal therefore represets 2/99. k
For most series, we will try to determie covergece or divergece, but we will ot be able to fid the sum. The geometric series is oe exceptio. Aother is the so called telescopig series. The followig example illustrates this case. Example. Fid the sum of the series + + + + + + 2 23 34 45 kk + Solutio. The techique of partial fractios shows that = kk + k ( k+ )
Thus we ca rewrite the th partial sum of the series as s = + + + = 2 23 2 + 2 3 + 3 4 + + + + = = + + The lim s = so + + + + + + = 2 23 34 45 kk k kk = + + =
Oce it was believed that if the terms of a series teded to 0, the the series coverged. This is false, as the followig example shows. =+ + + + k k = = 2 3 4 s =+ > + = 3 2 2 2 2 s = 4 s + + > 2 3 4 s + + > 2 4 4 2 s = 4 8 s + + + + > 4 5 6 7 8 s + + + + > 4 8 8 8 8 2 s > 6 5 2 ad i geeral s > +. 2 2 k+ = + + coverges. k 2 3 4 O the other had, ( )
Problem. Fid all values of x for which the series ( 3) k x =+ ( x 3) + ( x 3) 2 + ( x 3) 3 + 0 coverges, ad fid the sum of the series for those values of x. The series is geometric, ad the ratio is x 3. Thus we kow that it coverges whe x 3 <. This is whe < x 3 < or 2< x< 4. Whe it coverges, it coverges to = ( x 3) 4 x
Problem. Fid all values of x for which the series 3 k 3 9 27 = + + + 2 3 x k x x x coverges, ad fid the sum of the series for those values of x. The series is geometric, ad the ratio is 3/x. Thus we kow that it coverges whe 3/x <. This is whe < 3 <. This happes whe x/3 > or x/3 <, or x whe x > 3 or x < 3. 3 Whe it coverges, it coverges to x = 3 3 x 3 x