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)

- anonymous

- katieb

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- anonymous

It would be nice if you can also tell me which test is used to determine the convergence/divergence of the series

- 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

You familiar with this test?

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## More answers

- anonymous

yes

- anonymous

It works if the terms of your series are monotonic decreasing.

- 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

The only catch is that the terms of your sequence have to be monotone decreasing.

- anonymous

How about the fifth series

- anonymous

I'll look at it.

- anonymous

Also for the 4th one, you can't use the integral test right?

- 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

is that the nth term test?

- anonymous

might be - different names

- anonymous

Thanks :) How about the 4th one. Which test should i use? I just have difficult to decide what test should be used

- anonymous

I'll have a look.

- anonymous

sorry for bothering you :[

- anonymous

You're not bothering me :)

- anonymous

After a while, 6/(n^2-65) will just behave like 6/n^2 which diverges. Is that reasonable?

- anonymous

I'm just going through them now - had to do something.

- anonymous

:) NP

- 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

I have some advice - just give me a minute.

- anonymous

ok thnx (:

- 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

Thanks for your effort :)

- 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

Well, the denominator exists, but the quotient doesn't.

- anonymous

You comfortable with that, dichalao?

- anonymous

What you mean n=5 ->denominator does not exist?

- 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

(n^2-65) it is n^2-65

- anonymous

\[\frac{1}{0}= \infty\]You can remember it loosely as that.

- anonymous

and n can only be integers so ...it should be continious

- 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

This is what's happening with your series (summing the first seven terms):

- anonymous

\[-\frac{1}{24}-\frac{1}{21}-\frac{1}{16}-\frac{1}{9}+\frac{1}{0}+\frac{1}{11}+\frac{1}{24}+...\]

- anonymous

\[\lim_{c->0}\frac{1}{c}=\infty\]

- anonymous

Assuming the rest of the series is convergent, say to limit s, then you'd have, basically,\[s+\infty=\infty\]

- anonymous

It's therefore not convergent.

- anonymous

For what number of n you got how u got 1/0

- anonymous

you got 1/0

- anonymous

\[\frac{6}{n^2-25}\rightarrow \frac{6}{5^2-25}=\frac{6}{0}=6\times \frac{1}{0}\]

- anonymous

Sorry, was the 1 bit doing you in?

- 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

wait the original problem is (6/(n^2-65)) not 25

- anonymous

Oh God...I misread it...

- anonymous

It's convergent then

- anonymous

LOL time to get pair of glasses :P

- anonymous

At least you learnt something else!

- anonymous

Let me make sure properly, though.

- anonymous

Basically, it's asymptotically equivalent to 1/n^2 which is convergent (it's a p-series with p>1).

- anonymous

yea haha so 1,4 converges and the rest diverges?

- anonymous

so if i want to show how it is convergent. Which test should i use?

- anonymous

Number 4?

- anonymous

p-series test

- 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

Got It haha:) i might have one more problem would you mind help me out too

- anonymous

ok

- anonymous

you need to check the other problems i didn't explicitly cover here. i didn't look through all of them.

- anonymous

yea i did check the ones that you didn't cover :)

- 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

{1\over4}\sqrt{h} = sqrt(h)/4

- anonymous

I got part A..but i couldn't figure out part B

- anonymous

Can you type out the expressions in the equation editor?

- 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

Okay, I was thrown for a minute then remembered you're using imperial units.

- anonymous

Haha Take your time :)

- anonymous

Okay...I have an answer.

- anonymous

COOL!!

- 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

Are you expected to use mathematical induction to prove you have the correct series, or can it just be 'obvious'?

- 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

No just the answer is fine

- anonymous

I wrote the accumulated times up to 4 for you, then skipped to s_n. I'll scan it.

- anonymous

OH MY GOD you are the NICEST person i have ever met online :OOOOO

- anonymous

##### 1 Attachment

- 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

Well, according to my fried head. I'm *very* tired, so please double-check everything.

- anonymous

HAHA it is all good. Thanks for all your help. This really helps me out alot

- 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

Lhunardaien, did you get the link?

- anonymous

I am looking through it :)

- anonymous

i just looked now and thanx for link im looking now.

- anonymous

Lok,,can you please explain how to derive equation for the accumulated time for n step~

- anonymous

Yeah, look at the powers of r in relation to the subscript on s.

- 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

brb

- anonymous

\[1+r ^{1/2}+r+^{3/2}+.....) ->1(1-r^{n/2})/(1-r) this is the most confusing part

- 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

Here, a=1, and r is the r we're using anyway.

- anonymous

7/8, I think...

- anonymous

I input that equation , the system indicated that the equation is incorrect

- anonymous

hmm

- anonymous

is this a test?

- anonymous

This is a homework problem. (Webwork)

- anonymous

I think your analysis makes very much sense.

- anonymous

The accumulated sum is correct tho

- anonymous

I think it came down to a mismatch between the time-step wanted in the question and the one I used.

- anonymous

Ah...that confirms it.

- anonymous

The actual mathematics isn't wrong, it's just your instructor named the time-steps differently.

- anonymous

Siht

- anonymous

Sorry

- anonymous

Do you get another crack at it?

- anonymous

Did we start with n=1 right for the geometric series?

- anonymous

n=1 for *

- anonymous

Yes

- 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

http://mathworld.wolfram.com/GeometricSeries.html check out equation 8

- anonymous

Yeah, that's the case because the first term is r.

- anonymous

does that mean we use r^n instead of r^(n+1) for the partial sum

- anonymous

This is why I derive these things from first principles...because of this kind of crap that can happen. Bear with me.

- anonymous

So that would be (1-r^((n-2)/2)) /(1-r) instead of (1-r^(n/2)in this case?

- anonymous

Oh no...I know what happened - I should have derived it...

- 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

\[s_n=\frac{1}{4}\sqrt{h_0}\left[ 1+2r^{1/2}\frac{1-r^{\frac{n-1}{2}}}{1-r^{1/2}} \right]\]

- anonymous

the nth term is \[r^{n-2/2} \]
With the partial sum formula, how you got r^[(n-1)/2]

- anonymous

did you get 23.69s for the limiting time?

- anonymous

yea

- 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

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

how you got to\[ 1-r^{(n-1)/2}\]

- anonymous

okay

- 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

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

OH nice proof~!!!

- anonymous

You should get the same answer for your limiting time.

- anonymous

as n-> infinity , it should give the same result~

- anonymous

Yeah, p_n should approach \[\frac{1}{1-r^{1/2}}\]

- 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

if i plug in n=1,2,3, it should give answers to PART A as well?

- anonymous

try that

- anonymous

that would have been a useful check

- anonymous

YEP it works you are hero ~~!!!

- anonymous

oh good...

- anonymous

thank God

- anonymous

so does that mean you'll get it right for your homework?

- anonymous

i am typing in right now. Last attempt

- anonymous

sure you don't want to check some more...

- anonymous

?

- anonymous

I'm reeaaally tired and making mistakes at this hour.

- anonymous

It seems to work fine i tried it with n=1-4 ,and they all worked out fine

- anonymous

ok, if you're happy.

- anonymous

lol should i submit it..

- 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

it ..shouldn't work since (n-2)/2 will be negative when n =1

- anonymous

If you're happy with it. Is it worth much to your final grade?

- anonymous

i guess not

- anonymous

HIGH FIVE..IT IS CORRECT! What would i have done if this was on my midterm

- anonymous

AWESOME!!!!!!!!!!!!!!

- anonymous

i was nervous

- anonymous

LOL sorry for making you nervous~~haha

- 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

yea.. we should always try to derive equations

- anonymous

phew

- anonymous

are you good at physics as well?

- anonymous

i can sleep well

- anonymous

LOL i really should go to bed now..it is 5 in the morning here :)

- anonymous

omg

- anonymous

haha nice meeting you man YOU ARE GREAT !! Have a good one

- anonymous

no probs. good luck!

- anonymous

nice meeting you too

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