n 2 + 3n + 1 4n = n2 + 3n + 1 n n 2 = n + 1

Transcription

1 Ifiite Series Some Tests for Divergece d Covergece Divergece Test: If lim u or if the limit does ot exist, the series diverget EXAMPLE: Show tht the series diverges. = u = = lim = 4 u is = Compriso Test: Give two series of positive terms k d k k = k = k k for ll k elogig to the set of positive itegers, the if: is coverget to limit S, k k k = T S is diverget, so is k k k = k = k = such tht is lso coverget to limit T, where For exmple, is similr to the geometric series with commo rtio of d ifiite sum of 3 S = = 3. 3

2 Ech term i the first sequece is less th or equl to ech term of the secod so y the compriso test, the sequece coverges to sum less th 3. The Itegrl s the limit of sums Recll tht the re uder curve c e pproximted usig Riem Sums d exct vlue for the re c e foud y tkig the limit s the umer of rectgles used pproches ifiity, or s the chge i x pproches zero. i = f (c i ) Δx i where f ( + i Δx )gives the height of rectgle t its right edpoit d Δx i = f (x )dx is the width of ech rectgle i the itervl [,]. The defiite itegrl = F () F () i = f (c i ) Δx i EXAMPLE: For the fuctio f (x ) = e x d the prtitio of the itervl [,] y c i = i,i =,,3,... use suitle Riem sum to show tht lim + e + e + e 3 + e. ( ) = e f (c i ) Δx i = e c i i = e = + e + e + e 3 + e i = Hece, lim ( ) i = i = ( ) = e = + e + e + e 3 + e e

3 Improper Itegrls The defiite itegrl f (x )dx requires tht for the itervl, [ ], d must e rel umers d tht f must e cotiuous o the itervl. Itegrls tht do ot meet these requiremets re clled improper itegrls. Defiitio: A improper itegrl f (x )dx hs ifiity s oe or oth of its edpoits, [ ]. or cotis fiite umer of ifiite discotiuities i the itervl, Cosider the fuctio y = e x. Cosider the fuctio y = e x. It ppers tht the shded regio hs ifiite re, sice y = is horizotl symptote. We defie A() s the re uder the curve from = to =. A() = e x dx = e x = e + lim A() e + Hece, the re uder the curve from = to = is e x dx e x dx = More geerlly, if f(x) is cotiuous o the itervl [,) the f (x )dx f (x )dx

4 Covergece of improper itegrl If the limit exists, the improper itegrl coverges to the limitig vlue. If the limit fils to exist, the improper itegrl diverges. EXAMPLES: Evlute, if possile, the followig improper itegrls y discussig covergece. ) ) + x dx ) Diverges ecuse l[ x ] ( l) = ) This itegrl coverges ecuse + + rctx [ ] rct rct ( ) = π Itegrl Test for Covergece Torrecelli s Trumpet is oject with ifiite surfce re ut fiite volume.

5 It is defied y the series = = =. We c compre the fuctio x to Itegrl Test: Let f(x) e cotiuous, positive, d decresig fuctio. The the series f () + f () + f (3) + + f () + coverges if the improper itegrl f (x )dx coverges, d diverges if the itegrl diverges. EXAMPLE: Usig the itegrl test to determie if series coverges Let u = l + 3l3 + 4l4 + + l + First, exmie the ecessry coditio for covergece, tht lim l =. Sice f (x ) = x lx is icresig for ll vlues of x, the fuctio f (x ) = x lx is positive d decresig. x l u = lx,du = lim l l u du x lx l lu [ ] l ( l( l) l(l) ) = Sice the itegrl diverges, the sequece lso diverges.

6 The p- Series Test: The itegrl test c e used to determie covergece of series of the form where p is rel costt. Such series is clled p- series., p = coverges if p > p = diverges if p < p = diverges if p = p = Usig the itegrl test, Whe p =, lim Whe p >, lim x p dx lx [ ] (l ) = x p dp p + x p+ = p Usig Compriso Test Whe p <, p which we kow to diverge.

7 Rtio test for covergece: Let > for d lim + = L. The = coverges is L < d diverges if L >. If L =, the test is icoclusive. For exmple, if we pply the rtio test to the hrmoic series, we eed to evlute lim = However, we kow tht the hrmoic series diverges. Let us ow pply the rtio test to the series. = + + = ( + ) + = However, we kow this series coverges y the p- series test. ( + )

8 EXAMPLE: Determie the covergece or divergece of the followig series: ) = = = lim + = Sice L <, the series coverges + + ) =! =! + + ( + )!! = + ( + )!! Sice L <, the series coverges

9 Altertig Series Test ( ) k + k = k = ( ) If for ltertig series u k : k = u k + < u k for sufficietly lrge vlues of k lim u k = k the the series is coverget. Determie whether ( ) k + k k = is coverget. u k = k, u = k + k + < k + < for ll positive itegers k u k + < u k for ll positive itegers lim k u k k k = Therefore, y the Altertig Series Test ( ) k + k is coverget. k =

10 Oce we kow our ltertig series is coverget, we c go o to thik out pproximtio for the limit. If we were to stop, or tructe, this prtil sum t the th prtil sum, how ccurte would this sum e to the true limit, S? The distce etwee the th prtil sum d S, S S, kow s the tructio error, is less th the distce etwee the (+) th prtil sum d the th prtil sum S + S. S + = u + u + u u + u + S = u + u + u u S + S = u + ( ) If S = u k is the sum of ltertig series tht stisfies: k = u k + < u k for positive itegers, k lim k u k = the the error i tkig the first terms s pproximtio of the sum, S the tructio error is less th the solute vlue of the ( + ) th term. S S < u +

11 For exmple, tkig the first 9 terms of the ltertig hrmoic series will give error less th the th term for pproximtio of the limit. S S 9 < u = =. Tkig terms, S S 99 < u = =.