Applications: Series and Convergence

The layers are the square numbers, so the total is the sum of the first n squares. That sum has a closed formula, n(n+1)(2n+1)6, and the formula does both parts: it gives the twelve-layer total at once, and it lets the yard's limit be tested layer by layer without adding anything up again.

1. Count the pipes layer by layer. Layer 1 holds 1 pipe, layer 2 holds 2² = 4, layer 3 holds 3² = 9, and layer k holds k². A stack of n layers therefore holds ∑_k=1ⁿ k² pipes, the sum of the first n square numbers.
2. Use the closed formula rather than adding twelve terms: ∑_k=1ⁿ k² = n(n+1)(2n+1)6. With n = 12 the top is 12 × 13 × 25.
3. (a) 12 × 13 × 256 = 39006 = 650 pipes. Check by adding the twelve layers one at a time: 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144 = 650.
4. For part (b) the formula is quicker than a table. With n = 13 the stack holds 13 × 14 × 276 = 819 pipes, which is inside the limit. With n = 14 it holds 14 × 15 × 296 = 1015 pipes, which is over it.
5. (b) The greatest stack allowed is 13 layers. It holds 819 pipes, so it leaves 1000 − 819 = 181 pipes of the allowance unused. A fourteenth layer would need 14² = 196 more pipes, and only 181 are allowed.

answer(a) 650 pipes; (b) 13 layers, holding 819 pipes and leaving 181 of the 1000 unused

techniqueSums of Squares and Cubes

Adding fifteen fractions is possible but tells you nothing about the sixteenth hour. Writing each term as a difference of two neighboring fractions makes almost everything cancel, which gives the exact total after fifteen hours and, in the same line of working, the total after any number of hours at all.

1. Split the term into partial fractions. Since 1k − 1k+1 = 1k(k+1), the loss in hour k is 240k(k+1) = 240(1k − 1k+1) milliliters.
2. Write the fifteen hours out in that form: 240[(1 − 12) + (12 − 13) + … + (115 − 116)]. Every fraction after the first is subtracted once and added once, so the pairs cancel and only the two ends remain.
3. (a) What is left is 240(1 − 116) = 240 × 1516 = 225 mL.
4. The same canceling holds for any number of hours, so after n hours the total is 240(1 − 1n+1) mL. As n grows, 1n+1 tends to 0, so the total climbs to 240 mL without ever reaching it.
5. (b) Solve 240(1 − 1n+1) > 230, which gives 1n+1 < 10240 = 124, so n + 1 > 24 and n = 24. Check: after 23 hours the total is 240 × 2324 = 230 mL exactly, which has not passed 230; after 24 hours it is 240 × 2425 = 230.4 mL. The log is signed in hour 24, and the leak can never pass 240 mL.

answer(a) 225 mL; (b) in hour 24, and the leak can never pass 240 mL

techniqueThe Method of Differences

The trustee has run two different quantities together: the grant in a single year, which does settle, and the running total of all the grants, which is the sum of a series. The nth term test compares the terms with 0, and because they settle at 5 instead the series cannot converge. The year itself is then found by adding the grants until the running total passes the endowment.

1. Work out what one year's grant approaches. Divide the top and the bottom by n: 5nn+1 = 51 + 1n. As n grows, 1n tends to 0, so the grant tends to $5 thousand a year.
2. That is what the nth term test asks about. If ∑ a_n converges then a_n must tend to 0. Here a_n tends to 5, so the condition fails and the series of grants cannot converge.
3. (a) The total paid out diverges. The yearly grant settles at $5 thousand, but the running total of the grants grows without bound, and paying out nearly $5 thousand every year forever exhausts an endowment of any size.
4. For the year itself, add the grants one at a time. A rough guess is 100 ÷ 5 = 20 years, and the true answer is a little later because each grant is slightly under $5 thousand. The running totals are $86.8 thousand after 20 years, $96.3 thousand after 22 years and $101.1 thousand after 23 years.
5. (b) The total paid out first passes $100 thousand in year 23. The endowment is exhausted then, not never, and the trustee's plan fails in its twenty-third year.

answer(a) The grants tend to $5 thousand rather than to 0, so by the nth term test the total paid out diverges; (b) year 23

techniqueThe nth Term Test for Divergence

The bracket is the harmonic series, the p-series with p = 1. A p-series converges when p > 1 and diverges when p ≤ 1, so this one diverges and the reach has no ceiling at all. The number of planks for a stated reach is then a matter of adding terms until the total is large enough, and the count is large because the harmonic series grows so slowly.

1. (a) Put n = 4 into the bracket: 1 + 12 + 13 + 14 = 12 + 6 + 4 + 312 = 2512. The reach is 15 × 2512 = 37512 = 31.25 cm, which is already more than a whole plank's length past the edge.
2. Now look at the bracket as a series. It is ∑ 1n, the p-series with p = 1, which diverges: its partial sums grow without bound even though the terms shrink to 0.
3. Because the partial sums pass every number, some stack reaches 1 m. To find which, note that 1 m = 100 cm needs 1 + 12 + … + 1n = 10015 = 203 ≈ 6.667.
4. Add the terms one at a time until the total passes 203. At 440 planks the reach is 99.98 cm, just short; at 441 planks it is 100.01 cm.
5. (b) Yes, with 441 planks. The contrast is worth keeping in mind: if each plank added only 15k² cm, the p-series with p = 2 would converge and the whole stack could never reach 15 × π²6 ≈ 24.7 cm, however many planks were used.

answer(a) 31.25 cm; (b) yes, with 441 planks

techniqueThe p-Series · The nth Term Test for Divergence

Neither total can be added in closed form, and neither needs to be. The comparison test settles both: each term of the first series is below a term of a convergent p-series, and each term of the second is above a term of the harmonic series. Only the second question needs arithmetic, and then only adding terms until the total is large enough.

1. Compare P_n with a simpler term. For every n ≥ 1, 2n + 1 ≤ 3n and n³ + n ≥ n³, so P_n = 2n+1n³+n ≤ 3nn³ = 3n².
2. The series ∑ 3n² is three times the p-series with p = 2, which converges. By the comparison test a series of positive terms lying under the terms of a convergent series converges as well, so ∑ P_n converges.
3. (a) The total power is finite, and it is at most 3 ∑ 1n² = 3 × π²6 = π²2 ≈ 4.93 microwatts. Adding the first forty harmonics gives about 2.78 microwatts, comfortably inside that bound.
4. Turn to the second aerial. Here n² + n = n(n+1) and 2n + 1 = n + (n+1), so Q_n = 1n+1 + 1n, which is larger than 1n. The harmonic series diverges, so by the comparison test ∑ Q_n diverges and its total passes every number.
5. (b) Add the terms until the total passes 10. The first is Q₁ = 32 = 1.5 and the second is Q₂ = 56, and the totals reach 9.99 microwatts after 136 harmonics and 10.01 microwatts after 137. The aerial is switched off after 137 harmonics.

answer(a) Finite: every P_n ≤ 3n², so the total is at most π²2 ≈ 4.93 microwatts; (b) 137 harmonics

techniqueComparison Tests for Series · The p-Series

The moves alternate in direction and shrink in size, which is what the alternating series test is for. The test does two jobs here: it says the arm settles, and it hands over an error bound that costs nothing to work out, since the error after any number of steps is at most the size of the next move.

1. Check the alternating series test. The signs alternate; the sizes decrease, because 12n+1 < 12n for every n; and 12n tends to 0. All three conditions hold, so the series converges and the arm settles at a definite position.
2. Add the first four moves: 12 − 6 + 4 − 3 = 7 mm. After four steps the arm stands 7 mm from where it started.
3. For an alternating series that passes the test, the error after any number of terms is at most the size of the first term left out. Here that is the fifth move, 125 = 2.4 mm.
4. (a) The arm settles, and after four steps it is at 7 mm with at most 2.4 mm still to go. Consecutive partial sums straddle the settled value: 7 mm is below it and 7 + 2.4 = 9.4 mm is above it, so the arm settles between 7 mm and 9.4 mm. The exact value is 12 ln 2 = 8.32 mm.
5. (b) After n steps the bound is 12n+1 mm. Solve 12n+1 ≤ 0.05, which gives n + 1 ≥ 240 and n = 239 steps. Check: 12240 = 0.05 exactly, while 238 steps leave a bound of 12239 ≈ 0.0502 mm, which is more than 0.05.

answer(a) It settles; after four steps it stands at 7 mm, within 2.4 mm of the settled value, which lies between 7 mm and 9.4 mm and equals 12 ln 2 = 8.32 mm; (b) 239 steps

techniqueThe Alternating Series Test

A factorial is what the ratio test is best at: the ratio of one term to the one before cancels almost everything and leaves a simple expression in n. Here it leaves 3n+1, which tends to 0, so the series converges however large the average is. The ratio also shows how fast the terms die away, which is why so few of them are needed in part (b).

1. Write the ratio of one term to the one before it: a_n+1a_n = 3ⁿ⁺¹(n+1)! × n!3ⁿ = 3n+1, because (n+1)! = (n+1) × n! and every other factor cancels.
2. Let n grow. Then 3n+1 → 0, and 0 < 1, so by the ratio test ∑ 3ⁿn! converges. The same ratio says where the terms turn: it is above 1 at n = 0 and n = 1, equal to 1 at n = 2, and below 1 from n = 3 on, so the terms rise to a peak and then fall away quickly.
3. (a) The chances add to a finite number. In fact ∑_n=0^∞ 3ⁿn! = e³, so ∑ p_n = e⁻³ × e³ = 1, which is what a complete list of chances has to add to.
4. For part (b), being swamped means 6 or more calls, and it is shorter to find the chance of 5 or fewer and subtract. ∑_n=0⁵ 3ⁿn! = 1 + 3 + 4.5 + 4.5 + 3.375 + 2.025 = 18.4, so the chance of 5 or fewer calls is 18.4e⁻³ = 0.916.
5. (b) The chance of being swamped is 1 − 0.916 = 0.084, about one minute in twelve.

answer(a) The ratio is 3n+1, whose limit is 0, so the series converges and the chances add to 1; (b) about 0.084, roughly one minute in twelve

techniqueThe Ratio Test

A power series is trustworthy only inside its interval of convergence, so the first job is to find that interval: the ratio test gives the radius, and the two ends have to be tested separately because the ratio test says nothing there. Inside the interval the sum can be worked out; outside it there is no sum to quote.

1. Apply the ratio test to the sizes of the terms: |a_n+1a_n| = (n+1)|x|ⁿ⁺¹4ⁿ⁺¹ × 4ⁿn|x|ⁿ = n+1n × |x|4, and n+1n → 1, so the limit is |x|4.
2. The series converges while that limit is below 1, which is |x| < 4, so the radius of convergence is R = 4.
3. Test the two ends, which the ratio test leaves undecided. At x = 4 the terms are n × 4ⁿ4ⁿ = n, which do not tend to 0. At x = −4, a compression of 4 cm, they are (−1)ⁿ n, which do not tend to 0 either. By the nth term test both ends diverge.
4. (a) The radius is 4 and the interval of convergence is −4 < x < 4, with both ends excluded. The model says nothing at all about a stretch of 4 cm or more.
5. (b) At x = 2 the series is ∑_n=1^∞ n(12)ⁿ, and ∑_n=1^∞ nrⁿ = r(1−r)² for |r| < 1, so the energy is 12(12)² = 2 joules. At x = 5 the ratio is 54, which is above 1, so the terms grow without bound and the series diverges: the model gives nothing for a 5 cm extension, and a different model is needed there.

answer(a) R = 4, and the interval is −4 < x < 4 with both ends excluded; (b) 2 joules at 2 cm, and nothing at 5 cm, where the series diverges

techniqueRadius and Interval of Convergence · The Ratio Test · The nth Term Test for Divergence

The Maclaurin series for the sine is one of the standard five, so no derivatives need working out. What does need working out is how far a few terms can be from the true value, and the Lagrange bound answers that: every derivative of the sine is a sine or a cosine, so each is at most 1 in size and the bound is simply the next power over the next factorial.

1. Take the first three terms at x = 0.5: 0.5 − 0.5³6 + 0.5⁵120 = 0.5 − 0.0208333 + 0.0002604 = 0.4794271.
2. Bound the error. For a Taylor polynomial of degree n the Lagrange bound is M|x|ⁿ⁺¹(n+1)!, where M bounds the (n+1)th derivative. Every derivative of the sine is ±sin or ±cos, so M = 1 serves for all of them. Three terms reach degree 5, so the error is at most 0.5⁶6! = 0.015625720 ≈ 0.0000217.
3. (a) Six decimal places need the error below half a unit in the sixth place, which is 0.0000005. The bound 0.0000217 is far above that, so three terms cannot be trusted to 6 places, and in fact they are wrong there: the true value is 0.4794255, so the three-term estimate is out by about 0.0000015.
4. Try four terms, a polynomial of degree 7. The bound becomes 0.5⁸8! = 0.0039062540320 ≈ 0.000000097, which is below 0.0000005, so four terms are enough.
5. (b) Four terms give 0.5 − 0.5³6 + 0.5⁵120 − 0.5⁷5040 = 0.4794255, and sin 0.5 = 0.479426 to 6 decimal places.

answer(a) 0.4794271, with an error bound of 0.0000217, which is too large for 6 decimal places; (b) four terms, and sin 0.5 = 0.479426

techniqueThe Lagrange Error Bound · The Standard Maclaurin Series

A Maclaurin series is a Taylor series about zero, and zero degrees is no use here: the controller's information sits at 20 degrees. The polynomial is therefore written in powers of T − 20, the step away from the calibration point, and the Lagrange bound uses the largest third derivative anywhere between the calibration point and the temperature being asked about.

1. Write the Taylor polynomial of degree 2 about T = 20: S(T) ≈ S(20) + S'(20)(T − 20) + S''(20)2(T − 20)². The center is 20, not 0, so every power is a power of T − 20.
2. Put the stored numbers in, with 0.152 = 0.075: S(T) ≈ 30 − 1.5(T − 20) + 0.075(T − 20)².
3. (a) At T = 23 the step from the center is T − 20 = 3, so the estimate is 30 − 1.5 × 3 + 0.075 × 9 = 30 − 4.5 + 0.675 = 26.175 minutes.
4. For the error, the Lagrange bound after the degree 2 term is M|T − 20|³3!, where M bounds the third derivative between 20 and 23. Here S'''(T) = −3600T⁴, and 3600T⁴ is largest at the left end, so M = 360020⁴ = 3600160000 = 0.0225.
5. (b) The bound is 0.0225 × 3³6 = 0.60756 = 0.101 minutes, about 6 seconds, so the estimate of 26.175 minutes is within 0.101 minutes of the truth. It is: the exact time is 60023 = 26.087 minutes, and the estimate is out by 0.088 minutes.

answer(a) About 26.175 minutes; (b) within 0.101 minutes, about 6 seconds, and the exact time is 60023 = 26.087 minutes, so the estimate is out by 0.088 minutes

techniqueTaylor Series About a Point · The Lagrange Error Bound