Infinite Series

[warwick]

Will someone check my work and show me how to do #s 7.b and 13?

1) Find the sum of each series

3) Identify p and determine whether the series converges.

5) Apply the Divergence Test and state what it tells you about the series.

7) Confirm that the integral test is applicable and use it to determine whether the series converges.

9-24) Determine whether the series converges.

7.b Sum from k=1 to infinity of 1/(1 + 9k^2)

13. Sum from k=1 to infinity of 1/[(2k-1)^(1/3)]

http://www.freebibleproject.com/10-4a.jpg

 

[stapel]

Will someone check my work and show me how to do #s 7.b and 13?

1) Find the sum of each series.

3) Identify p and determine whether the series converges.

5) Apply the Divergence Test and state what it tells you about the series.

7) Confirm that the integral test is applicable and use it to determine whether the series converges.

9-24) Determine whether the series converges.

For 1, 3, 5, 7, and 9 - 24, what are the series?

7.b Sum from k=1 to infinity of 1/(1 + 9k^2)

13. Sum from k=1 to infinity of 1/[(2k-1)^(1/3)]

What work have you done on these?

http://www.freebibleproject.com/10-4a.jpg

I get a "forbidden" error. :shock:

Please reply, showing all of your work and reasoning. Thank you!

Eliz.

 

[pka]

7.b Sum from k=1 to infinity of 1/(1 + 9k^2)
13. Sum from k=1 to infinity of 1/[(2k-1)^(1/3)]

We cannot see your image.

But Sum from k=1 to infinity of 1/(1 + 9k^2)
Is no more that p=2, so it converges.

But #13 diverges because it is no more than p=(2/3).

 

[warwick]

Sum from k=1 to infinity of 1/(1 + 9k^2)
Is no more that p=2, so it converges.

But #13 diverges because it is no more than p=(2/3).

For Sum from k=1 to infinity of 1/(1 + 9k^2) I have to use the integral test.

For #13, I found the limit to be 0 which according to the DT it could converge or diverge. I can just call the 1/3 exponent "p"? I didn't know that.

 

[soroban]

Hello, warwick!

\(\displaystyle \L7)\;\;S\;=\;\sum^{\infty}_{k=1}\,\frac{1}{1\,+\,9k^2}\)

Since \(\displaystyle \,1\,+\,9k^2 \:\> \:k^2\), then: \(\displaystyle \frac{1}{1\,+\,9k^2} \:<\:\frac{1}{k^2\)

Hence: \(\displaystyle \L\:\sum^{\infty}_{k=1}\,\frac{1}{1\,+\,9k^2} \;< \;\sum^{\infty}_{k=1}\,\frac{1}{k^2}\;\) ... a convergent \(\displaystyle p\)-series \(\displaystyle \,(p\,=\,2)\)

Therefore, \(\displaystyle S\) converges.

\(\displaystyle \L13) \;\;S \;=\;\sum ^{\infty}_{k=1}\, \frac{1}{(2k\,-\,1)^{\frac{1}{3}}}\)

Since \(\displaystyle 2k\,-\,1 \:< \:2k\), then: \(\displaystyle \:\frac{1}{2k\,-\,1} \:> \frac{1}{2k}\)

Hence: \(\displaystyle \L\:\sum^{\infty}_{k=1}\,\frac{1}{(2k\,-\,1)^{\frac{1}{3}}} \;> \;\sum^{\infty}_{k=1}\,\frac{1}{(2k)^{\frac{1}{3}}} \;=\;\frac{1}{\sqrt[3]{2}}\sum^{\infty}_{k=1}\,\frac{1}{k^{\frac{1}{3}}}\) ... a divergent \(\displaystyle p\)-series.

Therefore, \(\displaystyle S\) diverges.

 

[warwick]

Hello, warwick!

\(\displaystyle \L7)\;\;S\;=\;\sum^{\infty}_{k=1}\,\frac{1}{1\,+\,9k^2}\)

Since \(\displaystyle \,1\,+\,9k^2 \:\> \:k^2\), then: \(\displaystyle \frac{1}{1\,+\,9k^2} \:<\:\frac{1}{k^2\)

Hence...

Thanks a lot.

Can you show me how to do #7 using the Integral Test?

 

[pka]

Soroban, Is that not what I had said.
Did you know that if a student copies your solution and uses it on a program called “BlackBoard” which is the most popular online teaching tool, in its last two versions has a feature that will identify if there is a match anywhere on the web. Therefore, the student will be flag as a cheater and it will be your fault. My question is what legal implications are there for such a board as this? I think that if I were you I would check my personal liability insurance. I carry $3,000,000 for just this reason. I may have written this before: but we have a well known southern boarding school in our community. Any student caught posting to such a site as this is expelled. The student’s parents loose an entire years cost, which can be in excess of $33,000.

 

[warwick]

Soroban, Is that not what I had said?

Am I missing something? The main reason why I post on this site is because I do not have a solutions manual. I honestly would really, really like to have one now.

 

[stapel]

Am I missing something?

The tutor was pointing out to that other poster that giving complete worked solutions not only does not (generally) help the student learn, but (nowadays) could cause the student, his family, and potentially this forum and its members a great deal of legal and financial difficulty. The other poster was being requested, again, please not to do students' homework for them.

For yourself, just make sure you understand the replies you receive, and that you write your answers in your own words, as your own work. :wink:

Eliz.

 

[warwick]

...just make sure you understand the replies you receive, and that you write your answers in your own words, as your own work.

Yes. For me, seeing something worked out does help.

For the #5.d answer, the book says "no information." Why?

\(\displaystyle \L\:\sum^{\infty}_{k=1}\,\frac{1}{k!} \;\)

For #7.b, I need to use the integral test. I guess I forgot how to integrate something of this form.

\(\displaystyle \L\:\sum^{\infty}_{k=1}\,\frac{1}{1\,+\,9k^2} \\)

#23

\(\displaystyle \L\:\sum^{\infty}_{k=5}\,\{7k}^{-1.01} \\)

S = lim as k approaches infinity of 7/k^1.01 = 0.

\(\displaystyle 7\L\:\sum^{\infty}_{k=5}\,\frac{1}{k^{1.01}} \\)

p > 1

So, it converges.

 

[Deleted member 4993]

For 7.b

What is the derivative of tan(ax)?

 

[warwick]

For 7.b What is the derivative of tan(ax)?

a sec^2(ax)

 

[o_O]

Whoops. Think arctan(x) was what was meant.

 

[pka]

Yes. For me, seeing something worked out does help.
For the #5.d answer, the book says "no information." Why?
\(\displaystyle \L\:\sum^{\infty}_{k=1}\,\frac{1}{k!} \;\)

For #7.b
\(\displaystyle \L\:\sum^{\infty}_{k=1}\,\frac{1}{1\,+\,9k^2} \\)

#23
\(\displaystyle \L\:\sum^{\infty}_{k=5}\,\{7k}^{-1.01} \\)
p > 1

Maybe seeing worked is a real hindrance to actually understanding the process?
Have you ever thought about it that way?

Consider these three problems above.
Each and every one of them is a basic comparison test problem.

All that is needed for the first is
\(\displaystyle \L k > 3\quad \Rightarrow \quad 2^k > k!\quad \Rightarrow \quad \frac{1}{{k!}} < \frac{1}{{2^k }}\)

Then the second
\(\displaystyle \L \frac{1}{{1 + 9k^2 }} < \frac{1}{{9k^2 }}\).

The last one is a p-series.

 

[Deleted member 4993]

For 7.b What is the derivative of tan(ax)?

a sec^2(ax)

Now look at derivative of \(\displaystyle tan^{-1}(ax)\)

 

[Deleted member 4993]

Whoops. Think arctan(x) was what was meant.

I was hoping the student will realize to use \(\displaystyle tan^{-1}(ax)\) with that prop - but did not happen.

 

[warwick]

I was hoping the student will realize to use \(\displaystyle tan^{-1}(ax)\) with that prop - but did not happen.

Sorry, guys. I just looked at it. I've been busy around the house. I now see the derivative of arctan x fits the form.

I assume the derivative of arctan ax will work?

 

[Deleted member 4993]

I assume the derivative of arctan ax will work? Make sure that it does....

 

[warwick]

I assume the derivative of arctan ax will work? Make sure that it does....

Well, I haven't been able to find the derivative of arc tan(ax), so I'll make a wild guess.

Using the Integral Test, the integral would be arc tan (3k) from 1 to infinity?

 

[warwick]

Well, keep in mind that sometimes you use techniques I haven't learned yet. Yesterday, we went over the Comparison, Ratio, and Root Tests.

Use any method to determine whether the series converges.

\(\displaystyle \L\:\sum^{\infty}_{k=1}\,\frac{k^{1/2}}{k^{3}+1} \\)