Calculus 2 - Examination II PRIOR INFORMATION: Although the exam is not supposed to be cummulative, you are still responsible for all of the information that we have covered in the course. As Improper Integration and the Integral Test, both involve integrals... Imagine that!... you should refresh your memory on the following: ) Direct Substitution. 2) Integration by Parts. (Section 7.) 3) Trig Integrals. (Section 7.2) 4) Trig Substitution. (Section 7.3) 5) Integration by Partial Fractions. (Section 7.4) As the trig functions and their inverses keep coming back, I suggest that you study their graphs. Since the first step to determining if a series is divergent is to find lim a n, it might be a very good idea to remember Indeterminate Forms and l Hôpital s Rule. IMPROPER INTEGRALS: There are two things to keep in mind here: ) I m not allowed to integrate something with one of the limits of integration as ±. 2) I m not allowed to integrate if the function has a discontinuity at any value between the limits of integration or even at the limits of integration. To integrate these functions, I have to break up the integral at any points of discontinuity. I also have to break up any integral that is improper at both limits of integration. I then look at each integral one at a time. If any of the integrals is divergent, then I m happy! I m done! The integral is DIVERGENT. To look at each integral, you have to substitute t for the problem number and take the limit as t goes to that number. For the values of x that are discontinuities, if you replace a lower limit with t, you should use a one-sided limit from the right, and if you repace an upper limit with t, you should use a one-sided limit from the left.
2 COMPARISON THEOREM: Suppose that f and g are continuous functions with f(x) g(x) 0 when x a. If a f(x) dx is convergent, then g(x) dx is ALSO convergent. a If a g(x) dx is divergent, then a f(x) dx is ALSO divergent. It s very important to remember the word ALSO. YOU CANNOT USE THE COMPARISON THEOREM TO SHOW THAT TWO INTE- GRALS ARE DIFFERENT. Either they both converge, or they both diverge. This Comparison Theorem and the Comparison Test (for SERIES) are summed up as follows: LARGER than DIVERGENT smaller than convergent IS DIVERGENT!!! is convergent... p- integrals : x p dx is convergent if p > and divergent if p. SEQUENCES: A sequence is just an ordered list of numbers. (, 3, 2, 4, 5, 7, 3, ) CONVERGENT/DIVERGENT SEQUENCES: A sequence is said to be CONVERGENT (or that it CONVERGES) if lim a n exists as a finite number. Otherwise the sequence is said to be DIVERGENT (or that it DIVERGES.) THM: If lim a n = L. lim f(x) = L and f(n) = a n when n is an integer, then x SQUEEZE THM: If a n b n c n for n n 0, and lim a n = lim c n = L, then lim b n = L. THM: If lim a n = 0, then lim a n = 0. THM: The sequence {r n } is convergent if < r and is divergent for all other values of r.
Increasing/Decreasing: A sequence is INCREASING if a n < a n+ and DECREASING if a n > a n+ for all n. If a sequence is strictly increasing or strictly decreasing it is called MONOTONIC. THM: Every bounded, monotonic sequence is convergent. SERIES: A SERIES is an infinite sum of a sequence. a n = a + a 2 + a 3 + + a n + a n+ + Partial Sums: s = a lim s n = s 2 = a + a 2 s 3 = a + a 2 + a 3 s n = a + a 2 + a 3 + + a n = a n n i= Relationship... It s important enough to rewrite: lim s n = If this limit exists, the sum is called CONVERGENT, otherwise the sum is called DIVERGENT. a n a i 3 GEOMETRIC SERIES: The geometric series ar n = a + ar + ar 2 + ar 3 + ( ) is convergent if r < and its sum is a r geometric series is divergent.. If r, the HARMONIC SERIES: The series GENT. n = + 2 + 3 + is DIVER-
4 THM: If the series a n is convergent, then lim a n = 0. DIVERGENCE TEST: If lim a n 0, then a n is DIVERGENT. INTEGRAL TEST (I.T.): Suppose f(n) = a n with f being ) continuous on [c, ), 2) positive on [c, ), and 3) decreasing on [c, ), where c is a positive integer. Then the series a n n=c and the integral CONVERGENT or BOTH DIVERGENT. c f(x) dx are either BOTH NOTE: Since the first million terms of the series do not factor into whether or not the series is convergent, c does not have to be. P-SERIES: The p-series p. n p is convergent if p > and divergent if DIRECT COMPARISON TEST (D.C.T.): Suppose that a n and b n are series with POSITIVE terms. ) If a n b n for all n and b n is convergent, then a n is ALSO CONVERGENT. 2) If a n b n for all n and b n is divergent, then b n is ALSO DIVERGENT. NOTE : integrals... This is the same as the comparison theorem for Greater than Divergent is Divergent, Less than Convergent is Convergent. NOTE 2: YOU MAY ONLY USE THIS TEST WHEN BOTH SERIES ARE POSITIVE!!!
5 LIMIT COMPARISON TEST (L.C.T.): Suppose that a n and b n are series with POSITIVE terms. ) If a n lim = c > 0, where c, then both series MATCH! b n 2) If lim converges. a n b n = 0 and b n converges, then a n ALSO diverges. a n 3) If lim = and b n diverges, then a n ALSO b n NOTE : If you choose the correct series only have the first situation. b n, you should NOTE 2: Do NOT use the limit comparison test when sin n and/or cos n is involved. Try using the direct comparison test instead. NOTE 3: YOU MAY ONLY USE THIS TEST WHEN BOTH SERIES ARE POSITIVE!!! ALTERNATING SERIES TEST (A.S.T.): If the alternating series ( ) n b n = b b 2 + b 3 or ( ) n b n = b + b 2 b 3 + satisfies three conditions: ) b n > 0 (Positive) 2) b n b n+ (Decreasing) 3) lim b n = 0 (Limit goes to 0.) then the series is convergent. NOTE : this test CANNOT be used to show that a series is DIVERGENT! If a series does not ( pass the third condition, use) the divergence test and the fact that lim ( )n b n does not exist to show that the series is divergent. NOTE 2: To make the terms alternate, we can use ( ) n, but we can also use trig... Example: cos(πn).
6 Alternating Series Estimation Theorem: The absolute value of the difference between a partial sum s n and the actual value of a convergent alternating series is less then the next term in the sequence (b n+ ). ABSOLUTE CONVERGENCE: A series a n is called absolutely convergent if the series of absolute values a n is convergent. *** THM ***: If a series is absolutely convergent, then is is convergent! CONDITIONALLY CONVERGENCE: A series a n is called conditionally convergent if an diverges, but a n converges. THE RATIO TEST: Part A: If lim a n+ a n = L <, then the series a n is absolutely convergent. STOP! Part B: If lim a n+ a n >, then the series a n is divergent. STOP! Part C: If lim a n+ a n =, then the test fails!!! Try another test. NOTE : There is no restrictions as to whether the series is positive or not as we are taking the absolute value of the terms. NOTE 2: If the test shows that the series is absolutely convergent, it is convergent also by a previous theorem, so STOP working on this problem. NOTE 3: If the test shows that the series is divergent, it will diverge by the divergence test, so STOP working on this problem. fail. NOTE 4: Do not use this test on rational expressions. It will NOTE 5: Do not use this test with trig functions unless the trig function is just alternating the sign +,, +,. NOTE 6: YOU MUST USE THIS TEST ON POWER SERIES to determine radius and interval of convergence.
7 THE ROOT TEST: Part A: If lim convergent. STOP! Part B: If lim n an = L <, then the series n an >, then the series a n a n is absolutely is divergent. STOP! Part C: If lim n an =, then the test fails!!! Try another test. NOTE : There is no restrictions as to whether the series is positive or not as we are taking the absolute value of the terms. NOTE 2: If the test shows that the series is absolutely convergent, it is convergent also by a previous theorem, so STOP working on this problem. NOTE 3: If the test shows that the series is divergent, it will diverge by the divergence test, so STOP working on this problem. TELESCOPING SERIES: Series that you can expand and then most of the terms will cancel. Usually you can identify them by one of the following three ways: ) A series with two parts, i.e. (an b n ) 2) A series that you can break up using partial fractions. 3) When they ask for the sum of a series that isn t geometric. NOTE: To evaluate this series, you must remember that a n = lim s n = lim (a + a 2 + a 3 + + a n ).
8 STRATEGIES!!!. If the series is of the form, it is a p-series, which we know np to be convergent if p > and divergent if p. 2. If the series has the form ar n or ar n, it is a geometric series, which converges if r < and diverges if r. 3. If the series has a form that is similar to a p -series or a geometric series, then one of the comparison tests should be considered. In particular, if a n is a rational function or algebraic function of n (involving roots of polynomials), then the series should be compared with a p-series. The value of p should be chosen by keeping only the highest powers of n in the numerator and denominator when they are multipied out. The comparison tests can only be used on series with positive terms. If we have a series with negative terms, we can apply the Comparison Test to a n to test for absolute convergence. If a series is absolutely convergent, it is also convergent. 4. If you can see at a glance that lim a n 0, then the Divergence Test should be used. 5. If the series is of the form ( ) n b n or a trig function that just alternates the sign, then the Alternating Series Test should be used. 6. Series that involve factorials or products of terms involving n should be tested with the Ratio Test. 7. If a n is of the form (b n ) n, then the Root Test might work. 8. Last, and certainly least, the Integral Test. Usually this involves natural logs or when you see a function and its derivative. This is the longest test and requires the most work, so try it last. This is a good guideline, but that is all it is. I don t use this exact method when I look at series, but it could help you out. Sometimes it is possible to determine the convergence/divergence of a series by more than one test. There are tons of problems that the integral test works for that one of the simpler comparison tests would work for as well.
9 POWER SERIES: or c n x n = c 0 + c x + c 2 x 2 + n=0 c n (x a) n n=0 THM: For a given power series, there are three possibilites: ) The series converges only when x = a. 2) The series converges for all x. 3) There is a possitive number R such that the series converges if x a < R and diverges if x a > R. In the third case, we have to find out what happens when x a = R and x a = R. R is called the radius of convergence. The interval of convergence is one of the following: (a R, a + R) (a R, a + R] [a R, a + R) or [a R, a + R] We use the other tests to determine whether the series is convergent for these two values of x by replacing x with the value of a R and then the value of a + R. If the series is convergent, we include those values of x. Often, we have to use two seperate tests to find out which of the endpoints gets included (if any).
0 First Sample Exam (-2). Omitted. 2. (FINAL) Determine the radius of convergence, r, and the interval (x 3) n of convergence of the power series n 4 n. 3. Omitted. 4. For which of the following series does the Ratio Test fail (i.e. give no result)? e /n n + 5 0 n () n 2 (2) 5 n (3) n! 5. Evaluate 6. (P-C) Given is equal to dx 2x or show divergent. ( cos ) n cos n + cos cos n +., show that the partial sum s n Using the definition of convergent series show whether the given series converges, and if it does, give its value. 7. (P-C) Show whether value if convergent. 2 e 3n converges or diverges, and find its n 8. (P-C) Show that the Integral Test can be applied to n 2 +. Use this test to determine whether this series converges or diverges. n + 9. (P-C) Show whether n e n converges or diverges. State explicitly which tests/theorems were used.
0. (P-C) part A: Show that ( ) n n n 2 + is convergent. part B: Find an estimate for the absolute value of the error if s 0 is used to approximate the sum.. (P-C) Determine whether converges conditionally, or diverges. n=2 ( ) n ln n n converges absolutely, e 2n 2. (P-C) Determine where 2e 2n + explicitly which tests/theorems were used. converges or diverges. State Second Sample Exam (3-25) 3. Determine the sum of the series 4 n + 3 n 5 n, provided it exists. 4. Determine whether each of the following series is convergent or divergent. ) 3 n 2) n 3 3) 3 n 5. Question omitted due to it being extremely STUPID. 6. Omitted. 7. Determine where each of the following series is convergent or divergent. ) cos(nπ) n 2) ( ) n n! 8. Does the series n=2 n ln n converge or diverge and why?
2 9. (P-C) Determine whether each SEQUENCE converges or diverges. If it converges, find the limit. { part A: {ln(3n + 4) ln(n)} part B: ( ) n sin } n part C: What may be concluded about the series application of the ratio test and why? part D: Does the series ln (2 + n ) 2 3n n 2 + 2 from an converge or diverge? Why? 20. (P-C) Determine whether the integral I = e is convergent or divergent. Evaluate it if it is convergent. x(ln x) 2 dx 2. (P-C) Test the series Justify your answer. 3 n 2 for convergence or divergence. 22. (P-C) Test the series for convergence or divergence. Justify your answer. 23. (P-C) Do #5 again. :) 3n 2 + 2n + 24. (P-C) Determine whether the series ( ) n is absolutely convergent, conditionally convergent, or divergent. Justify your ln(n + ) answer. 25. (P-C) Determine whether the series converges or diverges. Justify your answer. Third Sample Exam (26-37) 2 n n 3 n! 26. Determine if it if it is convergent. { } ln(2n) 27. Determine if ln(3n + ) convergent, find its limit. 0 e x dx is convergent or divergent. Evaluate + e2x is convergent or divergent. If it is
28. Determine the sum of the series 29. Determine the sum of the series n=2 ( n + n + 2 2 2n. n ) n +. 3 ) 30. Which of the following two series converge? ( ) n n 3. Suppose a n 2). Determine whether converges or diverges. If it is convergent, find its sum. e n n! (2n )! a n converges to A. Consider the series b n 32. (P-C) Partial sums s n = n the formula s n = n + 2 n. Part A: Find the sum of a n a i i= b n = of the series a n are given by or show that the series diverges. Part B: Find the formula for a n and simplify it. 33. (P-C) Approximate the following sum within 000 value using the Alternating Series Estimating Theorem. of the actual ( ) n n2 0 n 34. (P-C) Determine whether the following series converges or diverges. Justify your answer. n + 3 n5 + 2n
4 ANSWERS. (Omitted) 2. To determine interval and radius of convergence, I use the ratio test. (x 3) n+ lim (n + )4 n+ (x 3) n = lim n(x 3) (n + ) 4 = x 3 4 n 4 n I know from the ratio test that this series is convergent when the limit is less than, so the series I have is convergent when x 3 < which 4 means x 3 < 4. The radius of convergence for this series is 4. Solving x 3 < 4 for x, I get that 4 < x 3 < 4 and < x < 7. I now need to determine if the series is convergent when x = or when x = 7. ( 4) n When x =, our series is n 4 n = ( ) n which converges n by the Alternating Series Test. So is included in the interval of convergence. 4 n When x = 7, our series is n 4 n = which is divergent (Harmonic Series) and 7 is NOT included in the interval of n convergence. Hence the radius of convergence is 4 and the interval of convergence is [, 7).
5 3. Omitted. 4. It fails only for the first series a n+ lim, we get. a n e /n n 2 since when we take the 5. First I realize that this function has a discontinuity at 2. So I have to split it up. dx 2x = /2 dx 2x + dx /2 2x I only want to compute these integrals one at a time, because if any part diverges, the entire integral diverges. So... t /2 dx 2x = lim t dx t /2 2x = lim ln 2x t /2 2 = 2 ln 2t ln 3 which is divergent and I m done! 2 6. s n = a + a 2 + a 3 + + a n s n = (cos cos 2 ) + (cos 2 cos 3 ) + + (cos n cos n + ) s n = cos cos n + Since a n = lim s n = cos cos 0, we have that the telescoping series converges to (cos ).
6 7. When I see numbers raised to the n th power, I m hoping that it is a geometric series. Since there are no factorials, or polynomials that involve n, it is geometric. If there were factorials, or polynomial functions of n, I would use the ratio test. But this one is geometric: a = 2 e 3 and r =. This is probably easiest to see by writing out e3 the first two terms. 2 e 3n = 2 e 3 + 2 e 6 + Since r = ( ) e 3 <, it is convergent, and its sum is a ( ) r 2 2 e 3 /e 3 which simplifies to e 3. = 8. Let f(x) = x x 2 +. ) f is continuous on [, ). 2) f is positive on [, ). 3) f is decreasing on (, ). since f (x) = x2 (x 2 + ) 2 < 0 if x >. (SOMETIMES I will let you get away with saying that the function is decreasing... If it isn t completely obvious to 90% of the people in the class, I want you to prove that it is decreasing.) x dx x 2 + = lim t 2 ln(x2 + ) Since the integral diverges, Test. t = lim t n n 2 + [ 2 ln(t2 + ) 2 ln 2 ] = is also divergent by the Integral 9. I used the ratio test since I have a MIXTURE of polynomials and exponentials. The Direct Comparison Test and the Limit Comparison Test also work well here. n + 2 lim (n + )e n+ n + = lim n e n convergent by the Ratio Test. (n + 2)(n)e n (n + )e n+ (n + ) = e < so the series is
0. I recognize the alternating series, so I try that test. b n = n n 2 + ) b n is positive since n. 2) I have to show that it is decreasing. I do this the same way that I showed it to be decreasing in problem 8. 3) lim b n = 0 Therefore, by the Alternating Series Test, this series CONVERGES!!!. To test something for absolute convergence, I start by taking the ln n absolute value of the terms. This gives me n But I know that 0 < n < ln n n diverges by the divergence test. when n=2 7 n > 4, so THIS NEW series ALL THAT THIS SAYS ABOUT THE ORIGINAL SERIES is that it is NOT absolutely convergent. I then try the alternating series test to see if it is conditionally convergent... b n = ln n n ) b n is positive since n. 2) The sequence {b n } is decreasing since f (x) = ln x x 2 < 0 when x 4. 3) lim b n = 0 Therefore, by the Alternating Series Test, THIS series is convergent. Since the original series is convergent, but NOT absolutely convergent, it is called conditionally convergent. e 2n 2. Since lim 2e 2n + = 2 the Divergence Test. 0, we have that the series diverges by Whenever I am working with series, I keep in mind that the ONLY way for a series to be convergent is for the limit of the sequence of terms to be 0. This doesn t mean that it converges, but if the limit is something other than 0, the series MUST diverge! (Divergence Test.)
8 4 n + 3 n 3. 5 n = 4 n 5 n + 3 n 5 n These are both convergent geometric series. For the first one, both a and r are 4 5 and in the second one, they are both 3 5. Hence this series converges to 4 ( ) + 3 ( ) = 5 4/5 5 3/5 2 4. ) is a geometric series with r = < which is convergent. 3 2) is a p-series (p = 3 > ) so it is convergent. 3) is a p-series (p = 3 ) so it is divergent. 5. Answer omitted also. 6. Omitted again. 7. THIS IS NOT A partial-credit problem. I m not showing my work. You should try this problem on your own to verify my answer. ( ) n ) is the same as which is conditionally convergent. The n absolute value is which is divergent, but the original series is n convergent due to the Alternating Series Test. 2) converges absolutely by the Ratio Test. 8. THIS IS NOT A partial-credit problem. I m not showing my work. You should try this problem on your own to verify my answer. The series n=2 n ln n diverges by the Integral Test. 9. Part A: lim (ln(3n + 4) ln(n)) = lim converges to ln 3. ln 3n + 4 n = ln 3 so the sequence
9. (continued...) Part B: lim ( )n sin n = lim sin n = 0 Since the sequence of absolute values converges to 0, the sequence also converges to 0. ln Part C: Absolutely NOTHING as the test fails since the limit is. Part D: Since lim (2 + n ) 2 = ln 2 0, we have that series is divergent by the Divergence Test. e t e dx 20. x(ln x) 2 = lim dx t x(ln x) 2 = lim t Therefore the integral converges to. ln x t e [ = lim t ln t + ] = ln e ( 2. Since 3 n is a convergent geom. series r = ), and since 3 both series are positive, we can use the limit comparison test. lim Therefore, Test. 3 n 2 = lim 3 n 3 n 2 3 n 3 n 2 NOTE: The Direct Comparison test with = which means they MATCH!!! is also convergent by the Limit Comparison 9 will not work here!!! 3n 22. Since n 2 is a convergent p -series ( p = 2 ), and since both series are positive, we can use the limit comparison test. lim Therefore, is also convergent by the Limit Comparison Test. 3n 2 + 2n + n 2 n 2 = lim 3n 2 + 2n + = 3 3n 2 + 2n + 23. Another omission. which means they MATCH!
20 24. I m going to shift the index... ( ) n ln(n + ) = ( ) n ln n n=2 Now, we consider the absolute value... n=2 ln n. Since 0 < n <, and since is divergent (Harmonic Series), we have that is divergent by the Direct Comparison Test. ln n n=2 n ln n (THIS JUST TELLS US THAT THE ORIGINAL SERIES IS NOT AB- SOLUTELY CONVERGENT.) Use the Alternating Series Test: b n = ln n ) b n is positive since n 2. 2) {b n } is decreasing. 3) lim b n = 0 Therefore, by the Alternating Series Test, this series CONVERGES!!! Since the series is convergent, but NOT absolutely convergent, it is called conditionally convergent. 25. Since I have a mixture of factorials, exponential, and polynomial functions, I use the Ratio Test. lim 2 n+ (n + ) 3 (n + )! 2 n n 3 n! 2(n + ) 3 = lim n 3 (n + ) = 0 < We have that the series is absolutely convergent by the Ratio Test. 26. Converges to π 4 Use the substition u = ex. 27. Converges to. Rewrite in terms of x and use l Hôpitals Rule.
2 28. n=2 ( n + n + 2 n ) = lim n + (a 2 + a 3 + a 4 + + a n ) [( 3 = lim 4 2 ) ( 4 + 3 5 3 ) + + 4 ( 2 = lim 3 + n + ) n + 2 = 2 3 + = 3 ( n + n + 2 n )] n + 29. 2 2n = 4 + 6 + which is a geometric series a = 4 and r = 4. Since r = 4 convergent to ( ) = 4 /4 3. < it is 30. ) is convergent by the Alternating Series Test. 2) is convergent by the Ratio Test. 3. Since a n is convergent, lim a n = 0. Then lim b n = lim 0 so b n is divergent by the Divergence a n Test. 32. Part A: Part B: a n = lim s n = lim (n + 2 n ) = so the series diverges. a n = s n s n = n + 2 n (n 2 n+ ) = + 2 n
22 33. ( ) n n2 0 n = 0 + 4 00 9 000 + 6 0000 25 00000 + 25 Since is the first term < 00000 000 before it to use as our approximation. Our approximation: s 4 = 0 + 4 00 9 000 + 6 0000 Our maximum error is b 5 = 25 00000. we need to add all the terms 34. With problems that involve polynomials or rational functions or algebraic functions ( and the like), I would very much like to use the Limit Comparison Test. The highest power of n in the numerator (after multiplying it out) is n /2. the highest power of n in the denominator is n 5/3. I want to use the Limit Comparison Test with n/2 n 5/3 which reduces down to n. 7/6 Since I know that is a convergent p -series with (p = 7 n 7/6 6 > ), and since both series are positive, I can use the Limit Comparison Test to show that the other series MATCHES and will also be convergent. The nice thing is that I already know that the series will converge... I just have to prove it. n + 3 n5 + 2n n 7/6 n + lim = lim 3 n5 + 2n = n 7/6 Therefore both series match and our original series must also be convergent by the Limit Comparison Test.