Question

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)

Answer 1

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

Answer 2

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\[\lim_{c-> \infty}[2\log (x) - 2 \log (3 + x)|_1^c=\lim_{c-> \infty}[2\log \frac{x}{3+x}|_1^c\]\[=\lim_{c-> \infty}[2\log \frac{1}{3/x+1}|_1^c=2 \log 1 - 2 \log \frac{1}{4}=4 \log 2<\infty\]

Answer 3

You familiar with this test?

Answer 4

yes

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

How about the fifth series

Answer 9

I'll look at it.

Answer 10

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

Answer 11

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.

Answer 12

is that the nth term test?

Answer 13

might be - different names

Answer 14

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

Answer 15

I'll have a look.

Answer 16

sorry for bothering you :[

Answer 17

You're not bothering me :)

Answer 18

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

Answer 19

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

Answer 20

:) NP

Answer 21

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.

Answer 22

I have some advice - just give me a minute.

Answer 23

ok thnx (:

Answer 24

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...

Answer 25

Thanks for your effort :)

Answer 26

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.

Answer 27

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

Answer 28

You comfortable with that, dichalao?

Answer 29

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

Answer 30

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).

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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.

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

It's therefore not convergent.

Answer 40

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

Answer 41

you got 1/0

Answer 42

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

Answer 43

Sorry, was the 1 bit doing you in?

Answer 44

Since multiplicative constants don't do much, I just leave them out whenever it's appropriate. Sorry if that added to some confusion.

Answer 45

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

Answer 46

Oh God...I misread it...

Answer 47

It's convergent then

Answer 48

LOL time to get pair of glasses :P

Answer 49

At least you learnt something else!

Answer 50

Let me make sure properly, though.

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

Number 4?

Answer 55

p-series test

Answer 56

after acknowledging that, in the limit,\[\frac{6}{n^2-65} \]is asymptotically equivalent to \[\frac{6}{n^2}=6\frac{1}{n^2}\]

Answer 57

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

Answer 58

ok

Answer 59

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

Answer 60

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

Answer 61

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 =

Answer 62

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

Answer 63

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

Answer 64

Can you type out the expressions in the equation editor?

Answer 65

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 =

Answer 66

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

Answer 67

Haha Take your time :)

Answer 68

Okay...I have an answer.

Answer 69

COOL!!

Answer 70

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.

Answer 71

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

Answer 72

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...

Answer 73

No just the answer is fine

Answer 74

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

Answer 75

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

Answer 76

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.

Answer 77

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

Answer 78

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

Answer 79

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.

Answer 80

Lhunardaien, did you get the link?

Answer 81

I am looking through it :)

Answer 82

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

Answer 83

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

Answer 84

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

Answer 85

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

Answer 86

brb

Answer 87

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

Answer 88

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
\[\frac{a(1-r^{n+1})}{1-r}\]

Answer 89

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

Answer 90

7/8, I think...

Answer 91

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

Answer 92

hmm

Answer 93

is this a test?

Answer 94

This is a homework problem. (Webwork)

Answer 95

I think your analysis makes very much sense.

Answer 96

The accumulated sum is correct tho