MATH 142 Calculus 2 Written Homework 9 Solutions

downloaded from

Scholarfriends

https://scholarfriends.com Downloaded by: browsegrades@gmail.com | https://scholarfriends.com/singlePaper/304187/pennsylvania-state-university-math-calculus-2math-141-written-homework-9 Distribution of this document is illegal.

Written Homework 9

Math 141

Page 1 of 5

Question 1 Do the following series converge or diverge? Find the sum of any convergent series. For (a) and (b), you can assume the suggested pattern continues.

1 1 1 + − + ··· 2 4 8 (b) 1.1 + 1.12 + 1.13 + 1.14 + · · · ∞ X (c) 5(0.4n ) (a) 1 −

(d)

n=3 ∞ X

(−1)n 32n+2 n=1

Recall for geometric series with starting term a and ratio r we have ( a ∞ X = |r| < 1 n 2 ar = a + ar + ar + · · · .

1−r diverges |r| ≥ 1 n=0 1 (a) We see the ratio of each term compared to the previous term is r = − , and since | − 1/2| < 1, 2 we know this geometric series converges. We have

2 1 1 1 1 2 = 3 = 1+ − + − + ··· = .

1 2 2 3 1 − −2 2 (b) This geometric series diverges because r = 1.1 and |1.1| ≥ 1. (c) We have

∞ X

5(0.4n ) = 5(0.43 ) + 5(0.44 ) + 5(0.45 ) + · · · .

n=3

We see the ratio of each term compared to the previous term is r = 0.4, and since |0.4| < 1, we know this geometric series converges to 5(0.43 ) 5(0.43 ) 8 = = .

3 1 − 0.4 15 5 X (d) It doesn’t look quite like crn yet but if we do some basic exponent properties we’ll see that it is in fact a geometric series. ∞ X (−1)n n=1

3

= 2n+2 = =

∞ X (−1)n n=1 ∞ X

32n 32

(−1)n (32 )n · 9 n=1 ∞ X (−1)n

9 · 9n

n 1 1 = − 9 9 n=1

2 3 1 1 1 1 1 1 − + − − = + + ··· 9 9 9 9 9 9 n=1 ∞ X

1 We see the ratio of each term compared to the previous term is r = − and since | − 1/9| < 1, 9 we know this geometric series converges to

1 (1/9)(−1/9) 1 81

. = − 10 = − 90 1 − − 19 9

Written Homework 9

Math 141

Page 2 of 5

Question 2 A patient takes d mg of medicine every 8 hours. Every 8 hours, 80% of the medicine in the patient’s bloodstream is eliminated. The medicine has side effects if there is ever 15 mg or more in the patient’s blood, but there needs to be at least 2 mg in the patient’s blood for it to be effective. (a) What is the maximum safe value for d? (Hint: Consider An , the amount of medicine in the patient’s blood immediately after taking the nth dose, and think about what must be true about An to ensure a safe dosing schedule. Write out very carefully what A1 , A2 , A3 are before trying to find a pattern for An .) (b) What is the minimum effective value for d? (Hint: Consider Bn , the amount of medicine in the patient’s blood immediately before taking the nth dose.)

When a pill is taken the amount of medicine increases instantaneously by d mg. Then the body breaks all the medicine down by 80%, before the next dose of d occurs.

t

If we are worried about side effects (part (a)), then we need to watch that the spikes right after a dose stay below 15 mg. If we are worried about effectiveness, we need to make sure the dips stay above 2 mg. (a) We have A1 = d since A1 is immediately after the first dose, the medicine hasn’t been broken down at all yet.

However, immediately after the second dose we have A2 =

d + d(0.2) |{z} | {z } brand new second dose remnants of first dose after 8 hours

and we can deduce that A3 =

d + d(0.2) + d(0.2)2 . |{z} | {z } | {z } 3rd dose 2nd after 8 hrs 1st after 16 hours

The pattern seems to be An = d + d(0.2) + d(0.22 ) + · · · + d(0.2n−1 ),

Written Homework 9

Math 141

Page 3 of 5

to which we can apply the finite geometric sum formula An =

d − d(0.2n ) = 1.25d − 1.25d(0.2n ).

1 − 0.2

If we want a safe dosing schedule, we will want An ≤ 15 for all n — because if we’re worried about side effects we don’t want there to ever be too much right after a dose. We can accomplish this by requiring that lim An = 1.25d ≤ 15,

n→∞

and so d ≤ 12 is safe . (Here we used of course that since |0.2| < 1 that lim 0.2n = 0.) n→∞

(b) We have B1 = 0 since B1 is immediately before the first dose, no medicine has been taken yet. However, immediately before the second dose we have B2 =

d(0.2) | {z } remnants of first dose after 8 hours

B3 =

d(0.2) + d(0.2)2 . | {z } | {z } 2nd after 8 hrs 1st after 16 hours

and we can deduce that

The pattern seems to be Bn = d(0.2) + d(0.22 ) + · · · + d(0.2n−1 ), to which we can apply the finite geometric sum formula (remember starting term minus first missing, all divided by one minus the ratio) Bn =

d(0.2) − d(0.2n ) = 0.25d − 1.25d(0.2n ).

1 − 0.2

If we want an effective dosing schedule, we will want eventually that Bn > 2 for all n large enough — because if we’re worried about effectiveness we don’t want there to be too little in between doses, and since the medicine is broken down the lowest levels we need to watch out for are right before doses. We can accomplish this by requiring that lim Bn = 0.25d > 2,

n→∞

and so d > 8 is effective . (Here we used of course that since |0.2| < 1 that lim 0.2n = 0.) n→∞ Note: You could also have cleverly deduced that Bn = An − d

for all n

since the only thing that has happened in the instant between the two situations is that a new dose of d has been taken.

(Don’t worry about < vs ≤ for these problems, it’s not what we want the focus on this kind of problem to be.)

Written Homework 9

Math 141 Question 3 Determine whether the series

Page 4 of 5

∞ X

1 converges or diverges. Be sure to name any test(s) used n(ln n)3/2 n=2

as well as the key details.

This isn’t a p-series or geometric series, and the terms clearly go to zero (since n(ln n)3/2 → ∞ as n → ∞), the nth term divergence test doesn’t immediately give us the answer.

X 1 X1 You might be tempted to try limit comparison with something like or , but you’ll get n n2 things like

1 3/2 1 n(ln n)

lim = lim = 0, 1 n→∞ n→∞ (ln n)3/2 n so getting a finite nonzero limit won’t work out, since the (ln n)3/2 makes it too different from a p-series for the limit comparison test to think they’re similar.

Inequalities don’t work out in useful directions either — we know from L’Hospital’s rule that

1 ln n ln x H x lim = lim = lim =0 n→∞ n x→∞ x x→∞ 1 and so we must have ln n < n

for n large.

Then, (ln n)3/2 < n3/2 , n(ln n)5/2 < nn3/2 = n5/2 , 1 1 > 5/2 , n(ln n)3/2 n X 1 is a convergent p-series, (p = 5/2 > 1) the direct comparison test does not have n5/2 anything to say. (Walking ahead of somebody who doesn’t walk infinitely far could mean that you don’t either, or you could keep walking farther and farther!)

We don’t see anything like an or factorials, or f (n)n , and so we don’t see any ratio or root test hints either, and the series isn’t alternating. So the only thing left is the integral test — this is how you should think of the integral test, as a last resort when nothing else seems to work. Indeed, since 1 is continuous, positive, and decreasing for x ≥ 2, we can try the integral test and consider x(ln x)3/2 but since

Z ∞ 2

1 dx = lim t→∞ x(ln x)3/2

Z t

1 dx x(ln x)3/2 2 Z ln t 1 = lim du t→∞ ln 2 u3/2 ln t

u−1/2

= lim t→∞ − 1 2 ln 2

2 2 +√ = lim − √ t→∞ ln t ln 2 2 =√ , ln 2 hence the improper integral converges, and the series converges by the integral test . ∞ X

1 converges or diverges. Be sure to name any test(s) n · ln n · ln(ln n) n=3 used as well as the key details.

This isn’t a p-series or geometric series, and the terms clearly go to zero (since n ln n ln(ln n) → ∞ as n → ∞), the nth term divergence test doesn’t immediately give us the answer.

Question 4 Determine whether the series

Math 141

Written Homework 9

Page 5 of 5

You can try and explore some inequalities for Direct Comparison or a similar series for Limit Comparison, but nothing really works out. This is when you might want to start thinking about the integral test — we do see some hints for a u = ln x substitution. This is how you should think of the integral 1 test, as a last resort when nothing else seems to work. Indeed, since is continuous, x · ln x · ln(ln x) positive, and decreasing for x ≥ 3, we can try the integral test and consider

Z ∞ Z t 1 1 1 1 dx = lim dx, sub u = ln(ln x), du = dx, t→∞ x · ln x · ln(ln x) ln x x 2 2 x · ln x · ln(ln x) Z ln(ln t) 1 = lim du t→∞ ln(ln 2) u

= lim ln | ln(ln t)| − ln | ln(ln 2)| t→∞

= ∞, hence the improper integral diverges, and the series diverges by the integral test . As t → ∞, ln t → ∞, meaning ln(ln t) → ∞ too, meaning ln(ln(ln t)) → ∞ as well.