4.3. The Integral and Comparison Tests


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1 4.3. THE INTEGRAL AND COMPARISON TESTS The Itegral ad Compariso Tests The Itegral Test. Suppose f is a cotiuous, positive, decreasig fuctio o [, ), ad let a = f(). The the covergece or divergece of the series = a is the same as that of the itegral f(x) dx, i.e.: () If () If f(x) dx is coverget the f(x) dx is diverget the a is coverget. = a is diverget. The best way to see why the itegral test works is to compare the area uder the graph of y = f(x) betwee ad to the sum of the areas of rectagles of height f() placed alog itervals [, + ]. =.. y = f(x) y = f(x) y0.6 y0.6 f() f() f() f() 0. f(3) f(4) f(5) 0. f(3) f(4) f(5) f(6) x x Figure 4.3. From the graph we see that the followig iequality holds: + f(x) dx a f() + f(x) dx. i= The first iequality shows that if the itegral diverges so does the series. The secod iequality shows that if the itegral coverges the the same happes to the series. Example: Use the itegral test to prove that the harmoic series = / diverges.
2 4.3. THE INTEGRAL AND COMPARISON TESTS 9 Aswer: The covergece or divergece of the harmoic series is the same as that of the followig itegral: t [ ] t l x so it diverges. dx = lim x t dx = lim x t = lim t l t =, The pseries. The followig series is called pseries:. p = Its behavior is the same as that of the itegral we have see that it diverges. If p we have t [ ] x p t dx = lim dx = lim = lim xp t xp t p t dx. For p = x p t p p p. For 0 < p < the limit is ifiite, ad for p > it is zero so: The pseries = p is { coverget if p > diverget if p Compariso Test. Suppose that a ad b are series with positive terms ad suppose that a b for all. The () If b is coverget the a is coverget. () If a is diverget the b is diverget. coverges or diverges. Example: Determie whether the series = cos Aswer: We have = 0 < cos for all ad we kow that the pseries coverges. Hece by the compariso test, the give series also coverges (icidetally, its sum is π + π = , although we caot prove it here). 6
3 4.3. THE INTEGRAL AND COMPARISON TESTS The Limit Compariso Test. Suppose that a ad b are series with positive terms. If a lim = c, b where c is a fiite strictly positive umber, the either both series coverge or both diverge. Example: Determie whether the series diverges. = + 4 coverges or Aswer: We will use the limit compariso test with the harmoic series. We have = lim / + 4 / + 4 = lim + 4 = lim = lim + 4 = 4 =, so the give series has the same behavior as the harmoic series. Sice the harmoic series diverges, so does the give series Remaider Estimate for the Itegral Test. The differece betwee the sum s = = a of a coverget series ad its th partial sum s = i= a i is the remaider: R = s s = i=+ The same graphic used to see why the itegral test works allows us to estimate that remaider. Namely: If a coverges by the Itegral Test ad R = s s, the + f(x) dx R a i. f(x) dx
4 Equivaletly (addig s ): 4.3. THE INTEGRAL AND COMPARISON TESTS 94 s + + Example: Estimate f(x) dx s s + = f(x) dx to the third decimal place. 4 Aswer: We eed to reduce the remaider below , i.e., we eed to fid some such that dx < x4 We have [ x dx = ] = 4 3x 3 3, 3 hece 3 3 < > = , so we ca take = 9. So the sum of the 5 first terms of the give series coicides with the sum of the whole series up to the third decimal place: 9 i = i= From here we deduce that the actual sum s of the series is betwee = ad = , so we ca claim s.08. (The actual sum of the series is π4 = )
5 4.4. OTHER CONVERGENCE TESTS Other Covergece Tests Alteratig Series. A alteratig series is a series whose terms are alterately positive ad egative., for istace = ( ) The Alteratig Series Test. If the sequece of positive terms b verifies () b is decreasig. () lim b = 0 = the the alteratig series ( ) + b = b b + b 3 b 4 + coverges. = Example: The alteratig harmoic series = ( ) + coverges because / 0. (Its sum is l = ) Alteratig Series Estimatio Theorem. If s = = ( ) b is the sum of ad alteratig series verifyig that b is decreasig ad b 0, the the remaider of the series verifies: = R = s s b Absolute Covergece. A series = a is called absolutely coverget if the series of absolute values = a coverges. Absolute covergece implies covergece, i.e., if a series a is absolutely coverget, the it is coverget. The coverse is ot true i geeral. For istace, the alteratig harmoic series ( ) + = is coverget but it is ot absolutely coverget.
6 4.4. OTHER CONVERGENCE TESTS 96 Example: Determie whether the series cos is coverget or diverget. = Aswer: We see that the series of absolute values cos = is coverget by compariso with =, hece the give series is absolutely coverget, therefore it is coverget (its sum turs out to be /4 π/ + π /6 = , but the proof of this is beyod the scope of this otes) The Ratio Test. () If lim a + = L < the the series a is absolutely coverget. () If lim a a + a = = L > (icludig L = ) the the series is diverget. (3) If lim a + a = the the test is icoclusive (we do ot kow whether the series coverges or diverges). = a Example: Test the series for absolute covergece. = ( )! Aswer: We have: a + a = ( + )!/( + )+!/ = ( + ) = ( + hece by the Ratio Test the series is absolutely coverget. ) e <,
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