A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
For each of the following, determine whether the series is convergent or divergent.
1: sum x=1 to x=infinity 6/(n(n+3))
2: sum x=1 to x=infinity ln(n)/(2n)
3: sum x=1 to x=infinity 1/(3+n)
4: sum x=1 to infinity 6/(n^265)
5: sum x=1 to x=infinity (1+6^n)/(3+6^n)
anonymous
 5 years ago
For each of the following, determine whether the series is convergent or divergent. 1: sum x=1 to x=infinity 6/(n(n+3)) 2: sum x=1 to x=infinity ln(n)/(2n) 3: sum x=1 to x=infinity 1/(3+n) 4: sum x=1 to infinity 6/(n^265) 5: sum x=1 to x=infinity (1+6^n)/(3+6^n)

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It would be nice if you can also tell me which test is used to determine the convergence/divergence of the series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use the integral test on the first. It suffices to show that the integral of \[\lim_{c> \infty}\int\limits_{1}^{c}\frac{6}{x(x+3)}dx \]is convergent. When you integrate this out, you get\[\lim_{c> \infty}[2\log (x)  2 \log (3 + x)_1^c=\lim_{c> \infty}[2\log \frac{x}{3+x}_1^c\]\[=\lim_{c> \infty}[2\log \frac{1}{3/x+1}_1^c=2 \log 1  2 \log \frac{1}{4}=4 \log 2<\infty\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You familiar with this test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It works if the terms of your series are monotonic decreasing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have a feeling you could use this test for the others. The second one is divergent using the same test. Don't wast time  set it up and just let wolframalpha do the integral so you can check it out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The only catch is that the terms of your sequence have to be monotone decreasing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How about the fifth series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Also for the 4th one, you can't use the integral test right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah...you don't need to use the integral test for all of them. For the fifth, you can use the limit test. The test says that if the limit as n approaches infinity of a_n does not equal zero, the series will diverge. All you have to do is show the limit of (1+6^n)/(3+6^n) does not equal zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that the nth term test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0might be  different names

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks :) How about the 4th one. Which test should i use? I just have difficult to decide what test should be used

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry for bothering you :[

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're not bothering me :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0After a while, 6/(n^265) will just behave like 6/n^2 which diverges. Is that reasonable?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm just going through them now  had to do something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hi guys. i have midterm from same chapter. what is your advices about which method must be use for each equation. how can i decide it ? sorry for my poor english i hope you understand it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have some advice  just give me a minute.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know 4 diverges, and because of its form (polynomial fraction) I'm looking into the comparison test, or the limit comparison test. I'm trying to find an appropriate divergent sequence. I have to do this because the ratio test, etc. are inconclusive, and the integral (from the integral test) does not exist over all the domain we're interested in...so hang on...a little longer...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for your effort :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, sorry, got distracted again...but, I think I have your answer. Look at what happens when n=5 > the denominator does not exist. I was looking at this series and thinking, "Why is it not convergent if it's asymptotically equivalent to 1/n^2 (which is a pseries with p>1)...well, yeah, you're diving be zero at some point, which is undefined. If you exclude n=5, you'll get a convergent sum.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, the denominator exists, but the quotient doesn't.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You comfortable with that, dichalao?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What you mean n=5 >denominator does not exist?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't have a denominator equal to 0. When n is 5, the denominator is 5^225 = 25  25 = 0. I corrected myself above: the denominator exists, it's the quotient that doesn't (at this point).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(n^265) it is n^265

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{0}= \infty\]You can remember it loosely as that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and n can only be integers so ...it should be continious

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lhunardaien, here is a good rundown/strategy sheet. http://tutorial.math.lamar.edu/Classes/CalcII/SeriesStrategy.aspx If I find any more, I'll post them here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is what's happening with your series (summing the first seven terms):

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{24}\frac{1}{21}\frac{1}{16}\frac{1}{9}+\frac{1}{0}+\frac{1}{11}+\frac{1}{24}+...\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{c>0}\frac{1}{c}=\infty\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Assuming the rest of the series is convergent, say to limit s, then you'd have, basically,\[s+\infty=\infty\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's therefore not convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For what number of n you got how u got 1/0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{6}{n^225}\rightarrow \frac{6}{5^225}=\frac{6}{0}=6\times \frac{1}{0}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, was the 1 bit doing you in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since multiplicative constants don't do much, I just leave them out whenever it's appropriate. Sorry if that added to some confusion.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait the original problem is (6/(n^265)) not 25

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh God...I misread it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL time to get pair of glasses :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0At least you learnt something else!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me make sure properly, though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Basically, it's asymptotically equivalent to 1/n^2 which is convergent (it's a pseries with p>1).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea haha so 1,4 converges and the rest diverges?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if i want to show how it is convergent. Which test should i use?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0after acknowledging that, in the limit,\[\frac{6}{n^265} \]is asymptotically equivalent to \[\frac{6}{n^2}=6\frac{1}{n^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Got It haha:) i might have one more problem would you mind help me out too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need to check the other problems i didn't explicitly cover here. i didn't look through all of them.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i did check the ones that you didn't cover :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We might think that a ball that is dropped from a height of 10 feet and rebounds to a height 7/8 of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. This is not true! We examine this idea in this problem. A. Show that a ball dropped from a height h feet reaches the floor in {1\over4}\sqrt{h} seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times: time at first bounce = time at second bounce = time at third bounce = time at fourth bounce = B. How long, in seconds, has the ball been bouncing when it hits the floor for the nth time (find a closed form expression)? time at nth bounce = C. What is the total time that the ball bounces for? total time =

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0{1\over4}\sqrt{h} = sqrt(h)/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got part A..but i couldn't figure out part B

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you type out the expressions in the equation editor?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We might think that a ball that is dropped from a height of 10 feet and rebounds to a height 7/8 of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. A. Show that a ball dropped from a height h feet reaches the floor in \[{1\over4}\sqrt{h}\] seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times: time at first bounce = time at second bounce = time at third bounce = time at fourth bounce = B. How long, in seconds, has the ball been bouncing when it hits the floor for the nth time (find a closed form expression)? time at nth bounce = C. What is the total time that the ball bounces for? total time =

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, I was thrown for a minute then remembered you're using imperial units.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha Take your time :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...I have an answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When setting up your times, you have to keep in mind, for the first time, the ball falls only to the ground (from where it's dropped) BUT AFTER THAT, the time you calculate needs to be double at each step because it spends time going back up, and then back down to mark the next turn.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you expected to use mathematical induction to prove you have the correct series, or can it just be 'obvious'?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got part A,so i tried to come up with an expression ,,,i got I did that.. i got sqrt10/4+(sqrt10)(sqrt(7/8))^(n1)/2 for part B...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No just the answer is fine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I wrote the accumulated times up to 4 for you, then skipped to s_n. I'll scan it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH MY GOD you are the NICEST person i have ever met online :OOOOO

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0After the line for s_n, I jumped to the limit of an infinite sum. The formula on the second page is the accumulated time for the nth step.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, according to my fried head. I'm *very* tired, so please doublecheck everything.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0HAHA it is all good. Thanks for all your help. This really helps me out alot

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. Just check the accumulated sum. I usually derive the sums of geometric series on a casebycase basis, but this time, just pulled the formula. I don't think I'm wrong in my analysis, though, like I said, I'm fried, so the mathematics needs to be checked. Good luck.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lhunardaien, did you get the link?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am looking through it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just looked now and thanx for link im looking now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lok,,can you please explain how to derive equation for the accumulated time for n step~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, look at the powers of r in relation to the subscript on s.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When n=2, the highest power is 1/2 n=3, the highest power is 2/2 n=4, the highest power is 3/2 ... n=k, the highest power is (k1)/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1+r ^{1/2}+r+^{3/2}+.....) >1(1r^{n/2})/(1r) this is the most confusing part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, this is just applying the formula for the sum of a geometric series. For r<1, the sum of the first n terms of the series a+ar+ar^2+...+ar^n is \[\frac{a(1r^{n+1})}{1r}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here, a=1, and r is the r we're using anyway.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I input that equation , the system indicated that the equation is incorrect

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a homework problem. (Webwork)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think your analysis makes very much sense.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The accumulated sum is correct tho

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think it came down to a mismatch between the timestep wanted in the question and the one I used.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah...that confirms it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The actual mathematics isn't wrong, it's just your instructor named the timesteps differently.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you get another crack at it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did we start with n=1 right for the geometric series?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the accumulated sum (the limit) is correct, the analysis of the physics of the situation is correct, and there's been a difference in labeling.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://mathworld.wolfram.com/GeometricSeries.html check out equation 8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's the case because the first term is r.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does that mean we use r^n instead of r^(n+1) for the partial sum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is why I derive these things from first principles...because of this kind of crap that can happen. Bear with me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So that would be (1r^((n2)/2)) /(1r) instead of (1r^(n/2)in this case?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh no...I know what happened  I should have derived it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In the last formula, instead of \[\frac{1r^{n/2}}{1r}\]it should have been\[\frac{1r^{\frac{n1}{2}}}{1r^{1/2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[s_n=\frac{1}{4}\sqrt{h_0}\left[ 1+2r^{1/2}\frac{1r^{\frac{n1}{2}}}{1r^{1/2}} \right]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the nth term is \[r^{n2/2} \] With the partial sum formula, how you got r^[(n1)/2]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you get 23.69s for the limiting time?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the rate is r^(1/2) in this case.. so i think it suppose to be (1r^(n/2)/(1r^(1/2))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...I'm going to write out what I have\[s_n=\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}(1+r^{1/2}+r^{2/2}+...+r^{(n2)/2})\]\[=\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}.\frac{1r^{(n1)/2}}{1r^{1/2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how you got to\[ 1r^{(n1)/2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let the sum in brackets be equal to p_n,\[p_n=1+r^{1/2}+r^{2/2}+...+r^{(n2)/2}\]then multiply through by r^{1/2},\[r^{1/2}p_n=r^{1/2}+r^{2/2}+r^{3/2}+...+r^{(n1)/2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0subtract the second from the first,\[(1r^{1/2})p_n=1r^{(n1)/2} \rightarrow p_n=\frac{1r^{(n1)/2}}{1r^{1/2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should get the same answer for your limiting time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as n> infinity , it should give the same result~

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, p_n should approach \[\frac{1}{1r^{1/2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}.\frac{1r^{(n1)/2}}{1r^{1/2}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if i plug in n=1,2,3, it should give answers to PART A as well?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that would have been a useful check

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YEP it works you are hero ~~!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so does that mean you'll get it right for your homework?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am typing in right now. Last attempt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure you don't want to check some more...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm reeaaally tired and making mistakes at this hour.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It seems to work fine i tried it with n=14 ,and they all worked out fine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol should i submit it..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0check the one you put in last time that was wrong  check to see you DON'T get the right answers to part A.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it ..shouldn't work since (n2)/2 will be negative when n =1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you're happy with it. Is it worth much to your final grade?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0HIGH FIVE..IT IS CORRECT! What would i have done if this was on my midterm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0AWESOME!!!!!!!!!!!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL sorry for making you nervous~~haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My advice  learn to derive the results for geometric sums instead of using a formula. See what happened when I just used a formula?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea.. we should always try to derive equations

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you good at physics as well?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL i really should go to bed now..it is 5 in the morning here :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha nice meeting you man YOU ARE GREAT !! Have a good one
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.