anonymous
  • anonymous
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^2-65) 5: sum x=1 to x=infinity (1+6^n)/(3+6^n)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
It would be nice if you can also tell me which test is used to determine the convergence/divergence of the series
anonymous
  • anonymous
You 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
  • anonymous
You familiar with this test?

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anonymous
  • anonymous
yes
anonymous
  • anonymous
It works if the terms of your series are monotonic decreasing.
anonymous
  • anonymous
I 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
  • anonymous
The only catch is that the terms of your sequence have to be monotone decreasing.
anonymous
  • anonymous
How about the fifth series
anonymous
  • anonymous
I'll look at it.
anonymous
  • anonymous
Also for the 4th one, you can't use the integral test right?
anonymous
  • anonymous
Yeah...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
  • anonymous
is that the nth term test?
anonymous
  • anonymous
might be - different names
anonymous
  • anonymous
Thanks :) How about the 4th one. Which test should i use? I just have difficult to decide what test should be used
anonymous
  • anonymous
I'll have a look.
anonymous
  • anonymous
sorry for bothering you :[
anonymous
  • anonymous
You're not bothering me :)
anonymous
  • anonymous
After a while, 6/(n^2-65) will just behave like 6/n^2 which diverges. Is that reasonable?
anonymous
  • anonymous
I'm just going through them now - had to do something.
anonymous
  • anonymous
:) NP
anonymous
  • anonymous
hi 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
  • anonymous
I have some advice - just give me a minute.
anonymous
  • anonymous
ok thnx (:
anonymous
  • anonymous
I 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
  • anonymous
Thanks for your effort :)
anonymous
  • anonymous
Okay, 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 p-series 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
  • anonymous
Well, the denominator exists, but the quotient doesn't.
anonymous
  • anonymous
You comfortable with that, dichalao?
anonymous
  • anonymous
What you mean n=5 ->denominator does not exist?
anonymous
  • anonymous
You can't have a denominator equal to 0. When n is 5, the denominator is 5^2-25 = 25 - 25 = 0. I corrected myself above: the denominator exists, it's the quotient that doesn't (at this point).
anonymous
  • anonymous
(n^2-65) it is n^2-65
anonymous
  • anonymous
\[\frac{1}{0}= \infty\]You can remember it loosely as that.
anonymous
  • anonymous
and n can only be integers so ...it should be continious
anonymous
  • anonymous
Lhunardaien, here is a good run-down/strategy sheet. http://tutorial.math.lamar.edu/Classes/CalcII/SeriesStrategy.aspx If I find any more, I'll post them here.
anonymous
  • anonymous
This is what's happening with your series (summing the first seven terms):
anonymous
  • anonymous
\[-\frac{1}{24}-\frac{1}{21}-\frac{1}{16}-\frac{1}{9}+\frac{1}{0}+\frac{1}{11}+\frac{1}{24}+...\]
anonymous
  • anonymous
\[\lim_{c->0}\frac{1}{c}=\infty\]
anonymous
  • anonymous
Assuming the rest of the series is convergent, say to limit s, then you'd have, basically,\[s+\infty=\infty\]
anonymous
  • anonymous
It's therefore not convergent.
anonymous
  • anonymous
For what number of n you got how u got 1/0
anonymous
  • anonymous
you got 1/0
anonymous
  • anonymous
\[\frac{6}{n^2-25}\rightarrow \frac{6}{5^2-25}=\frac{6}{0}=6\times \frac{1}{0}\]
anonymous
  • anonymous
Sorry, was the 1 bit doing you in?
anonymous
  • anonymous
Since multiplicative constants don't do much, I just leave them out whenever it's appropriate. Sorry if that added to some confusion.
anonymous
  • anonymous
wait the original problem is (6/(n^2-65)) not 25
anonymous
  • anonymous
Oh God...I misread it...
anonymous
  • anonymous
It's convergent then
anonymous
  • anonymous
LOL time to get pair of glasses :P
anonymous
  • anonymous
At least you learnt something else!
anonymous
  • anonymous
Let me make sure properly, though.
anonymous
  • anonymous
Basically, it's asymptotically equivalent to 1/n^2 which is convergent (it's a p-series with p>1).
anonymous
  • anonymous
yea haha so 1,4 converges and the rest diverges?
anonymous
  • anonymous
so if i want to show how it is convergent. Which test should i use?
anonymous
  • anonymous
Number 4?
anonymous
  • anonymous
p-series test
anonymous
  • anonymous
after acknowledging that, in the limit,\[\frac{6}{n^2-65} \]is asymptotically equivalent to \[\frac{6}{n^2}=6\frac{1}{n^2}\]
anonymous
  • anonymous
Got It haha:) i might have one more problem would you mind help me out too
anonymous
  • anonymous
ok
anonymous
  • anonymous
you need to check the other problems i didn't explicitly cover here. i didn't look through all of them.
anonymous
  • anonymous
yea i did check the ones that you didn't cover :)
anonymous
  • anonymous
We 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
  • anonymous
{1\over4}\sqrt{h} = sqrt(h)/4
anonymous
  • anonymous
I got part A..but i couldn't figure out part B
anonymous
  • anonymous
Can you type out the expressions in the equation editor?
anonymous
  • anonymous
We 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
  • anonymous
Okay, I was thrown for a minute then remembered you're using imperial units.
anonymous
  • anonymous
Haha Take your time :)
anonymous
  • anonymous
Okay...I have an answer.
anonymous
  • anonymous
COOL!!
anonymous
  • anonymous
When 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
  • anonymous
Are you expected to use mathematical induction to prove you have the correct series, or can it just be 'obvious'?
anonymous
  • anonymous
I 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))^(n-1)/2 for part B...
anonymous
  • anonymous
No just the answer is fine
anonymous
  • anonymous
I wrote the accumulated times up to 4 for you, then skipped to s_n. I'll scan it.
anonymous
  • anonymous
OH MY GOD you are the NICEST person i have ever met online :OOOOO
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
After 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
  • anonymous
Well, according to my fried head. I'm *very* tired, so please double-check everything.
anonymous
  • anonymous
HAHA it is all good. Thanks for all your help. This really helps me out alot
anonymous
  • anonymous
Okay. Just check the accumulated sum. I usually derive the sums of geometric series on a case-by-case 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
  • anonymous
Lhunardaien, did you get the link?
anonymous
  • anonymous
I am looking through it :)
anonymous
  • anonymous
i just looked now and thanx for link im looking now.
anonymous
  • anonymous
Lok,,can you please explain how to derive equation for the accumulated time for n step~
anonymous
  • anonymous
Yeah, look at the powers of r in relation to the subscript on s.
anonymous
  • anonymous
When 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 (k-1)/2
anonymous
  • anonymous
brb
anonymous
  • anonymous
\[1+r ^{1/2}+r+^{3/2}+.....) ->1(1-r^{n/2})/(1-r) this is the most confusing part
anonymous
  • anonymous
Okay, 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 \[\frac{a(1-r^{n+1})}{1-r}\]
anonymous
  • anonymous
Here, a=1, and r is the r we're using anyway.
anonymous
  • anonymous
7/8, I think...
anonymous
  • anonymous
I input that equation , the system indicated that the equation is incorrect
anonymous
  • anonymous
hmm
anonymous
  • anonymous
is this a test?
anonymous
  • anonymous
This is a homework problem. (Webwork)
anonymous
  • anonymous
I think your analysis makes very much sense.
anonymous
  • anonymous
The accumulated sum is correct tho
anonymous
  • anonymous
I think it came down to a mismatch between the time-step wanted in the question and the one I used.
anonymous
  • anonymous
Ah...that confirms it.
anonymous
  • anonymous
The actual mathematics isn't wrong, it's just your instructor named the time-steps differently.
anonymous
  • anonymous
Siht
anonymous
  • anonymous
Sorry
anonymous
  • anonymous
Do you get another crack at it?
anonymous
  • anonymous
Did we start with n=1 right for the geometric series?
anonymous
  • anonymous
n=1 for *
anonymous
  • anonymous
Yes
anonymous
  • anonymous
If 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
  • anonymous
http://mathworld.wolfram.com/GeometricSeries.html check out equation 8
anonymous
  • anonymous
Yeah, that's the case because the first term is r.
anonymous
  • anonymous
does that mean we use r^n instead of r^(n+1) for the partial sum
anonymous
  • anonymous
This is why I derive these things from first principles...because of this kind of crap that can happen. Bear with me.
anonymous
  • anonymous
So that would be (1-r^((n-2)/2)) /(1-r) instead of (1-r^(n/2)in this case?
anonymous
  • anonymous
Oh no...I know what happened - I should have derived it...
anonymous
  • anonymous
In the last formula, instead of \[\frac{1-r^{n/2}}{1-r}\]it should have been\[\frac{1-r^{\frac{n-1}{2}}}{1-r^{1/2}}\]
anonymous
  • anonymous
\[s_n=\frac{1}{4}\sqrt{h_0}\left[ 1+2r^{1/2}\frac{1-r^{\frac{n-1}{2}}}{1-r^{1/2}} \right]\]
anonymous
  • anonymous
the nth term is \[r^{n-2/2} \] With the partial sum formula, how you got r^[(n-1)/2]
anonymous
  • anonymous
did you get 23.69s for the limiting time?
anonymous
  • anonymous
yea
anonymous
  • anonymous
the rate is r^(1/2) in this case.. so i think it suppose to be (1-r^(n/2)/(1-r^(1/2))
anonymous
  • anonymous
Okay...I'm going to write out what I have\[s_n=\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}(1+r^{1/2}+r^{2/2}+...+r^{(n-2)/2})\]\[=\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}.\frac{1-r^{(n-1)/2}}{1-r^{1/2}}\]
anonymous
  • anonymous
how you got to\[ 1-r^{(n-1)/2}\]
anonymous
  • anonymous
okay
anonymous
  • anonymous
let the sum in brackets be equal to p_n,\[p_n=1+r^{1/2}+r^{2/2}+...+r^{(n-2)/2}\]then multiply through by r^{1/2},\[r^{1/2}p_n=r^{1/2}+r^{2/2}+r^{3/2}+...+r^{(n-1)/2}\]
anonymous
  • anonymous
subtract the second from the first,\[(1-r^{1/2})p_n=1-r^{(n-1)/2} \rightarrow p_n=\frac{1-r^{(n-1)/2}}{1-r^{1/2}}\]
anonymous
  • anonymous
OH nice proof~!!!
anonymous
  • anonymous
You should get the same answer for your limiting time.
anonymous
  • anonymous
as n-> infinity , it should give the same result~
anonymous
  • anonymous
Yeah, p_n should approach \[\frac{1}{1-r^{1/2}}\]
anonymous
  • anonymous
\[\frac{1}{4}\sqrt{h_0}+\frac{1}{2}\sqrt{h_0}r^{1/2}.\frac{1-r^{(n-1)/2}}{1-r^{1/2}}\]
anonymous
  • anonymous
if i plug in n=1,2,3, it should give answers to PART A as well?
anonymous
  • anonymous
try that
anonymous
  • anonymous
that would have been a useful check
anonymous
  • anonymous
YEP it works you are hero ~~!!!
anonymous
  • anonymous
oh good...
anonymous
  • anonymous
thank God
anonymous
  • anonymous
so does that mean you'll get it right for your homework?
anonymous
  • anonymous
i am typing in right now. Last attempt
anonymous
  • anonymous
sure you don't want to check some more...
anonymous
  • anonymous
?
anonymous
  • anonymous
I'm reeaaally tired and making mistakes at this hour.
anonymous
  • anonymous
It seems to work fine i tried it with n=1-4 ,and they all worked out fine
anonymous
  • anonymous
ok, if you're happy.
anonymous
  • anonymous
lol should i submit it..
anonymous
  • anonymous
check the one you put in last time that was wrong - check to see you DON'T get the right answers to part A.
anonymous
  • anonymous
it ..shouldn't work since (n-2)/2 will be negative when n =1
anonymous
  • anonymous
If you're happy with it. Is it worth much to your final grade?
anonymous
  • anonymous
i guess not
anonymous
  • anonymous
HIGH FIVE..IT IS CORRECT! What would i have done if this was on my midterm
anonymous
  • anonymous
AWESOME!!!!!!!!!!!!!!
anonymous
  • anonymous
i was nervous
anonymous
  • anonymous
LOL sorry for making you nervous~~haha
anonymous
  • anonymous
My advice - learn to derive the results for geometric sums instead of using a formula. See what happened when I just used a formula?
anonymous
  • anonymous
yea.. we should always try to derive equations
anonymous
  • anonymous
phew
anonymous
  • anonymous
are you good at physics as well?
anonymous
  • anonymous
i can sleep well
anonymous
  • anonymous
LOL i really should go to bed now..it is 5 in the morning here :)
anonymous
  • anonymous
omg
anonymous
  • anonymous
haha nice meeting you man YOU ARE GREAT !! Have a good one
anonymous
  • anonymous
no probs. good luck!
anonymous
  • anonymous
nice meeting you too

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