Project: The Harmonic Series, the Integral Test

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1 . Let s start with the sequence a n = n. Does this sequence converge or diverge? Explain. To determine if a sequence converges, we just take a limit: lim n n sequence converges to 0. = 0. The. Now consider this infinite series (called the harmonic series): (a) Write this series in summation notation n= n () Now consider this series numerically. Calculate the following partial sums: n = n= n = 3 =.5 n= 3 n.833 n= 4 n.083 n= 0 n= 00 n= 000 n= n.99 n 5.87 n 7.48 (c) Recall that convergence of an infinite series is determined y taking the limit of the partial sums. In other words, m n m n n= If the limit is finite, we say the infinite series converges, otherwise we say it diverges. Using what you calculated in part() of this prolem, make a conjecture aout whether or not the harmonic series converges. The numers don t appear to e settling down yet, which makes me suspicious that it won t converge. But the numers are growing slowly, so I m suspicious that the total won t grow without ound (diverge to infinity). I m going to guess that it diverges, ut it s really just a guess. n=

2 3. The next part of the project introduces the concept of the Integral Test to show a series diverges. (a) Every series can e depicted graphically. Write down a sum that gives the area of the shaded region elow. How does this sum relate to the harmonic series? / /3 /4 / The width of each rectangle is unit, the heights are,, 3, etc. So the total area is , the harmonic series. () The advantage of representing a series this way is that it can e compared to an improper integral. On the aove graph, carefully draw the function f(x) = x. (c) How does the harmonic series compare to the improper integral n x dx? n= The improper integral is contained in the region that represents the harmonic series. The sum of the harmonic series is igger than the area representing the improper integral. (d) Does the improper integral dx converge or diverge? Calculate it, as a review of x improper integrals. It diverges. dx x x dx ln x (ln ln ) ln = (e) What can you conclude aout the convergence of the harmonic series Since the series is larger than the integral, and the integral diverges, the series must also diverge. n= n?

3 4. In the previous prolem we compared an infinite series to an improper integral to show divergence of the infinite series. By shifting to the left where we draw the rectangles, we can compare an infinite series to an improper integral to show convergence of the series. (a) Write down a sum that gives the area of the shaded region elow. How does this sum relate to the series n? n= /4 /9 /6 / The width of each rectangle is unit, the heights are, 4, 9, etc. So the total area is , the series n. n= () On the aove graph, carefully draw the function f(x) =. x (c) How does the series compare to the improper integral n n= x dx? The improper integral contains the region that represents the series. The sum of the series is smaller than the area representing the improper integral. (d) Does the improper integral improper integrals. It converges. dx x dx converge or diverge? Calculate it, as a review of x x dx x dx x [( ) ( () )] [ ] + = 0 + = (e) What can you conclude aout the convergence of the series series n= n? 3 n=, and thus aout the n

4 The integral is larger than the first sum, and since the integral converges, the first sum must also converge. The second sum is only larger than the first sum, and since the first sum converges, the second sum must also converge. 5. In prolems 3 and 4 we compared infinite series to improper integrals in order to make conclusions aout the convergence of divergence of the infinite series. Here is the general result: The Integral Test: Suppose f is a continuous, positive, decreasing function on [, ) and let a n = f(n). Then the series a n is convergent if and only if the improper integral n= f(x) dx converges. In other words: if we know if we know f(x) dx converges, then we know f(x) dx diverges, then we know a n converges. n= a n diverges. n= 6. Why do you think we need f(x) to e decreasing? Think aout what might go wrong with the geometrical argument if f(x) isn t decreasing. If f(x) matches a n at the integers, ut isn t forced to e decreasing, then inetween the integers f(x) could swoop aove the rectangles whose areas represent a n. Then our argument that a n is larger than the improper integral wouldn t hold. Consequently, in some cases f(x) dx could diverge even if a n converges. 7. Now we ll apply the Integral Test in an example. Determine whether or diverges y following these steps: n= n + converges a n = n + ; so let f(x) = x + f(x) is decreasing ecause: (Hint: find f (x)) f (x) = x(x + ), which is negative. So f(x) is decreasing. Check that f(x) is positive: Yes, ecause x is positive, + x is positive, and is positive. + x 4

5 We ve fulfilled the hypotheses of the integral test, so we can conclude that x dx either oth converge or oth diverge. + Integrate to determine whether x dx converges or diverges: + n= n + and x dx + x + dx arctan x (arctan arctan ) arctan π 4 = π π 4 = π 4 x + dx converges, and therefore n + 8. Use the integral test to determine whether n= n ln n n= also converges. converges or diverges. Let f(x) =. Since n >, oth n and ln n are positive, and so f(x) is x ln x positive. f ( + ln n) (x) = (n ln n) < 0, so f(x) is decreasing. The integral test applies, so now we ll check for convergence of the improper integral. We conclude dx x ln x x ln x dx (sustitute u = ln x, and du = x dx) ln ln u du ln u ln ln ln ln ln ln = dx diverges, and thus x ln x x dx diverges. + 5