Series: Infinite Sums
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- Laurence Morrison
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1 Series: Ifiite Sums Series are a way to mae sese of certai types of ifiitely log sums. We will eed to be able to do this if we are to attai our goal of approximatig trascedetal fuctios by usig ifiite degree polyomials. But before we try to add together a ifiite umber of polyomials, we first explore what it meas to add a ifiite umber of umbers. Here s the issue: We ow how to add two umbers: a a 2. Usig associativity (ad paretheses) we ca add three umbers a (a 2 a 3 ) four umbers a (a 2 (a 3 a )) or eve umbers a (a 2 (a 3 (a ( (a a )... )))). But where would we start (or ed) whe tryig to add a ifiite umber of terms? Ad does the sum add up to a fiite umber or ot? Sice all we ow how to do is add a fiite umber of terms, we will have to use fiite additio ad limits to mae sese of the process. Itroductio to Series OK, eough of this fiite stuff. What we wat to do is add up the terms of a ifiite sequece {a } =. More precisely, give a sequece {a } =, we ca form the ifiite sum a a 2 a 3 a = a which is called a ifiite series or more simply just a series. Ca we do this? Here are several examples. (a) (c) (d) = 2 3 =. The sum is clearly ot fiite; the series diverges. = 2 3. Do these terms add up to a fiite sum? 2 = 2. Do these terms add up to a fiite sum? ( 2) = 2. Do these terms add up to a fiite sum?
2 math 3 ifiite series, part i: itroductio 2 DEFINITION 3.. To fid the sum of a ifiite series a we form the sequece of partial sums that are ofte deoted by S. S = a S 2 = a a 2 S 3 = a a 2 a 3. S = a a 2 a 3 a = a (S is called the th partial sum of the series) If the sequece of partial sums {S } = has a limit a limit L (coverges), we say that the series coverges to L ad we write: or just Otherwise the series diverges. a = lim a = L. a = lim S = L EXAMPLE 3.. Here s a simple example. Fid the sum of the series if it exists. 2 = 2 6, SOLUTION. We first determie each partial sum ad the rewrite it i a more coveiet form. S = 2 = 2 S 2 = 2 = 3 = S 2 = 2 = 7 =. S = 2 2 = 2 So the sequece of partial sums is {S } = = { 2 } = ad lim S = lim 2 = 0 =, where we have used Theorem 3.2 to evaluate the limit. I other words, Pretty cool! 2 =. EXAMPLE 3.2. Here s a aother fu example. Fid the sum of the series it exists. SOLUTION. Usig partial fractios (chec this) 2 = 2 if ( ) ( = ) ( 2 2 ) ( 3 3 ) ( )
3 3 Notice that most of the terms cacel out. The sum collapses ad we see that ( S = ) ( 2 2 ) ( 3 3 ) ( ) ( ) =. Such a sum is called a telescopig sum. We are left with oly the first ad last terms i the partial sum. This time lim S = lim = 0 =. I other words, 2 =. ( YOU TRY IT 3.. Try this telescopig sum. Fid the sum of the series ) if it 2 exists. This time there will be a few more terms that do ot cacel. See if you ca figure it out. EXAMPLE 3.3 (Partial Fractios). Here s aother example that uses partial fractios. Fid the sum of the series =0 2 if it exists. 3 2 SOLUTION. Sice the degree of the umerator is smaller tha the degree of the deomiator a d sice the deomiator factors ito liear factors, we ca write = ( )( 2) = A Solvig we get: ad Subtractig (3.) from (3.2) gives B 2 Puttig A = i (3.) maes B =. So we see that A 2A B B =. ( )( 2) s : 0 = A B. (3.) costats : = 2A B. (3.2) = 2. (Chec that this is correct!) This meas that = = =0 which is aother telescopig series. This time ( S = ) ( 2 2 ) ( 3 3 = A. (3.3) ( ) 2 ) ( ) = 2 2. So I other words, Wow! lim S = lim 2 = 0 =. = =. EXAMPLE 3. (Telescopig). Here s a more complicated example that uses partial ( ) fractios. Fid the sum of the series l if it exists.
4 math 3 ifiite series, part i: itroductio SOLUTION. We ca use a log property to rewrite the partial sum as ( ) S = l = l( ) l = (l 2 l ) (l 3 l 2) (l l 3) [l( ) l ] = l l. Therefore ad the series diverges. lim S = lim l l = (diverges) ( ) l EXAMPLE 3. (Partial Fractios). Here s a more complicated example that uses partial fractios. Fid the sum of the series =0 2 if it exists. 3 SOLUTION. Sice the degree of the umerator is smaller tha the degree of the deomiator a d sice the deomiator factors ito liear factors, we ca write 2 3 = ( )( 3) = A Solvig we get: ad Subtractig (3.) from (3.) gives B 3 A 3A B B =. ( )( 3) s : 0 = A B. (3.) costats : = 3A B. (3.) = 2A. (3.6) Puttig A = i (3.) maes B =. So we see that This meas that =0 which is aother telescopig series. ( ) ( ) S = 3 2 So I other words, S = = = =0 ( ( ) 3 ( ) ( ) 3 ) ( 2 lim S = lim = 6. =0 2 3 = 6. ) ( 3 YOU TRY IT 3.2 (Partial fractios). Here are two others that are similar to the last example i that they use partial fractios. See if you ca solve them. Fid the sums of these series if they exist. (a) = =0 2 3 ) Aswers: (a) ; 3 2.
5 Geometric Series Geometric series are amog the simpler with which to wor. We will see that we ca determie which oes coverge ad what their limits are fairly easily. DEFINITION 3.2. A geometric series is a series that has the form costat ad r is a real umber. ar, where a is a real YOU TRY IT 3.3. Here are a few examples. Idetify a ad r i each. (a) (c) Aswers: (a) ; /2; (c) /3. Determiig the sum of a geometric series by comparig the th partial sum S with rs. We fid: ar is relatively simple. We begi S = a ar ar 2 ar 3 ar (3.7) rs = ar ar 2 ar 3 ar ar (3.) So subtractig (3.) from (3.7) we obtai or S rs = a ar ( r)s = a( r. So We ow from the Key Limit Theorem 3.2 that lim r = S = a( r ). (3.9) r 0 if r < if r = diverges otherwise. Thus, puttig (3.9) ad (3.0) together we fid lim S a( r = ) a r if r < lim = r diverges otherwise. So we have proved THEOREM 3. (Geometric Series Test). If r <, the the geometric series ad ar = a r. If r, the the geometric series ar diverges. (3.0) ar coverges
6 math 3 ifiite series, part i: itroductio 6 EXAMPLE 3.6. Here are some examples that get progressively more complex. ( ) 2 (a) Fid the sum of the series if it exists. SOLUTION. I this example a = ad r = 2 ad r <. So by Theorem 3. the series coverges to = 2 3. Fid the sum of the series ( ) 6 if it exists. 7 SOLUTION. I this example a = ad r = 7 6 ad r <. So by Theorem 3. the series coverges to = 2 6 = 2. 7 (c) (d) Fid the sum of the series ( ) 3 2 if it exists. 2 SOLUTION. I this example a = 2 ad r = 2 3. Sice r >, by Theorem 3. the series diverges. Fid the sum of the series ( ) 2 if it exists. 2 SOLUTION. Before we ca apply the Geometric Series Test, we have to adjust the power. Notice that we ca rewrite the series usig the th power usig ( ) 2 = 2 ( ) 2 ( ) = 2 2 ( ). 2 Now a = ad r = 2 ad r <. So by Theorem 3. the series coverges to ( 2 ) = 3 2 = 6. SOLUTION. Alterative Method. Aother way that we ca approach this problem is to write out the first few terms of the series ad idetify a ad r. ( ) 2 = 2 }{{} a }{{} ar }{{} 6 ar 2 32 }{{} ar 3. Now a = ad the ratio of a term to the previous oe is r = 2 ad r <. So by Theorem 3. the series coverges to ( 2 ) = 3 2 = 6. I fid this easier!
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