2 n = n=1 a n is convergent and we let. i=1
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1 Lecture 3 : Series So far our defiitio of a sum of umbers applies oly to addig a fiite set of umbers. We ca exted this to a defiitio of a sum of a ifiite set of umbers i much the same way as we exteded our otio of the defiite itegral to a improper itegral over a ifiite iterval. We call this ifiite sum a series = Defiitio Give a series a = a + a + a , we let s deote its th partial sum s = a + a + + a. If the sequece {s } is coverget ad lim s = S, the we say that the series a is coverget ad we let a = lim i= a = lim s = S. The umber S is called the sum of the series. Otherwise the series is called diverget. Fid the partial sums s, s, s 3,..., s of the series Fid the sum of this seies. Does the series coverge? Recall that = (+). Does the series coverge? The geometric series is coverget if r < ad its sum is Geometric Series ar = a + ar + ar + ar = If r, the geometric series is diverget. Proof If r =, ad the series diverges. If r =, a r r <.. ar = a + a + a + a +, s = a, lim s = ar = a a+a a+, s = a if is odd ad s = 0 if is eve, lim s = does ot exist
2 ad the series diverges. If r, we have Thus we get rs = s = a + ar + ar + ar ar ar + ar + ar ar + ar s rs = a ar or s ( r) = a( r ) or s = a( r ) ( r) If < r <, we saw i the sectio o sequeces, that lim r = 0 ad thus lim s = a ( r) givig us the desired result. If r >, the lim r a( r does ot exist ad hece lim s = lim ) ( r) the series does ot coverge. does ot exist. Thus Fid the sum of the series Fid the sum of the series ( ) 0 4 = = Fid the sum of the series =4 3 For which values of x does the series x coverge? Write the umbers = 0. 6 ad.5 =.5 as fractios. Telescopig Series These are series of the form similar to f() f( + ). Because of the large amout of cacellatio, they are relatively easy to sum. Show that the series coverges. k= k + 7k + = (k + 3) (k + 4) k=
3 Show that the series coverges. k= k k+ Harmoic Series The followig series, kow as the harmoic series, diverges: We ca see this if we look at a subsequece of partial sums: {s }. k= s = Similarly we get s = + = 3 s 4 = + [ ] > + [ ] = 4 s 8 = + [ ] [ [ > s 4 + 8] ] + > + = 5 s > + ad lim s > lim + =. Hece the harmoic series diverges. (You will see a easier proof i the ext sectio. ) Note that covergece or divergece is uaffected by addig or deletig a fiite umber of terms at the begiig of the series. ad 0 k=50 is diverget k is coverget. Divergece Test Theorem If a series i= a is coverget, the lim a = 0. Warig The coverse is ot true, we may have a series where lim a diverget. For example, the harmoic series. Proof Suppose the series i= a is coverget with sum S. Sice a = s s ad lim s = lim s = S we have lim a = lim s lim s = S S = 0. = 0 ad the series i 3
4 This gives us a Test for Divergece: If lim a does ot exist or if lim a 0, the i= a is diverget. If lim a = 0 the test is icoclusive. Test the followig series for divergece with the above test: Note that if lim a = 0, this test is icoclusive ad the series may diverge or coverge. Properties of Series The followig properties of series follow from the correspodig laws of limits: Suppose a ad b are coverget series, the the series (a + b ), (a b ) ad ca also coverge. We have ca = c a, (a + b ) = a + b, (a b ) = a b. Sum the followig series: k= k + 7k + 3 k. Because both of these series coverge we ca break it ito the differece of two series to sum it. k= from our previous calculatios. k + 7k + 3 = k k= = /4 k + 7k + k= 3 k 4
5 Puzzle: At o a rubber bad. A at starts at oe ed of a oe meter rubber bad, placed coveietly at x = 0 o the x axis. Iitially the other ed of the rubber bad is at x =. Each secod the at walks cm. At the ed of each secod Mike, who likes teasig ats, stretches the rubber bad by oe meter. (Note the poit at which the at is at moves whe the bad is stretched.). Will the at ever reach the ed of the rubber bad? Hit calculate the proportio of the distace covered by the at after secod, secods 3 secods,..., secods, ad derive your aswer from the sum of the series. 5
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