Question

Calc 2 Question on convergence and divergence...

Answer 1

\[\sum_{n =0}^{\infty}\frac{ n! }{ 2*5*8*...*(3n+2) }\]

Answer 2

I'm fairly certain it is the ratio test. But I'm not sure if I'm allowed to ignore the 2*5*8... in the denominator or not and just rewrite it as\[\sum_{n=0}^{\infty}\frac{ n! }{ 3n+2 }\]

Answer 3

well, you have it written out correct that way that's true, can you continue with the ratio test?

Answer 4

Not sure.. I'm getting \[\lim_{n \rightarrow \infty}\frac{ a_{n+1} }{a_{n} }=\frac{ n!(n+1) }{ 3n+3 }\frac{ 3n+2 }{ n! }=\frac{(3n+2)(n+1)}{3n+3}=\frac{3n^{2}+5n+2}{3n+3}\] (I know technically the limit should be in front of all but for brevity's sake I omitted it) But then carrying out l'Hopital's rule I get the lim = infinity. Would that just be divergence then since infinity > 1? Or do you need a finite number for the ratio test to tell you anything?

Answer 5

Well done, but please check again, I get:
\[\Large \frac{(n+1)!}{(3(n+1)+2)}\cdot \frac{3n+2}{n!}=\frac{(n+1)(3n+2)}{3n+5} \]
Which would be:
\[\Large \frac{3n^2+5n+2}{3n+5} \]
only a minor difference anyway, so from here continue, this is easy to check which limit it approaches as \(n \to \infty\) just divide the quotient by \(n\)

Answer 6

So, I should get \[\frac{3n+5+\frac{2}{n}}{3+\frac{5}{n}}\]? Essentially the same, is the ratio test valid for when the limit = infinity?

Answer 7

Well this is not the most rigorous attempt you can make, maybe you'd rather choose the path by taking the long hand division. But anyhow, you can tell from there that it will not converge, the order of the numerator is bigger than the order of the denominator. Or do you require a specific value?

Answer 8

just convergence or divergence. Although, I still feel like I'm going about this wrong because the original function appears that we wouldn't be able to ignore the 2*5*8... on the bottom.

Answer 9

or here maybe if you want to use L'Hopital's rule:

\[\Large \lim_{n \to \infty}\frac{3n^2+4n+2}{3n+5}=\lim_{n \to \infty} \frac{6n+4}{3}=\frac{\infty}{3}=\infty \]

Answer 10

so it does diverge = not converge.

Answer 11

wait
no, I think it will converge.

Answer 12

Because the series is really \[a_{1} =\frac{1}{2}\]\[a_{n} = \frac{n*a_{n-1}}{3n+2}\]\[a_{n+1}=\frac{(n+1)*a_{n}}{3n+5}\]

Answer 13

I have my doubts that this function will converge :-) But I'd sure like to see your attempt, from my knowledge of time complexity in algorithms n! is one of the fastest growing functions out there, so basically, no matter what you plug into the denominator, the numerator will always grow faster.

Answer 14

http://www.wolframalpha.com/input/?i=Sum_0%5Einfty+%28n%21%29%2F%283n%2B5%29

Answer 15

So, \[\frac{a_{n+1}}{a_{n}}=\frac{\frac{(n+1)a_{n}}{3n+5}}{\frac{a_{n}}{1}}\]the a(n)s would cancel leaving you with \[\frac{n+1}{3n+5}=\frac{1+\frac{1}{n}}{3+\frac{5}{n}}\] and the limit of that would go to 1/3 and thus converge no?

Answer 16

But you said yourself that \[\Large a_n=\frac{na_{n-1}}{3n+2} \]

so a depends upon the value we choose of n, what you did above seems to me like you're just dividing by a specific n value of a, which I haven't seen being done before, it's after all like dividing by a number. If you plugin the above, things wont cancel out nicely.

Answer 17

no no no. Because since each new term in the sequence a(n) is \[\frac{n}{3n+2}\]multiplied by the previous term in the sequence, or a(n-1) then a(n+1) would be \[\frac{n+1}{3n+5}\] multiplied by a(n). Then when you divide a(n+1) by a(n) you would be dividing a(n) by a(n) leaving you with \[\frac{n+1}{3n+5}\]

Answer 18

Hmm I see where you're trying to get, to be honest, I haven't seen such an approach yet, I am used to work straight up with the sum. From the sum it is pretty obvious for me that it will diverge, as in agreement with wolframalpha (you can check your answer there too)

Answer 19

Except, what you put in to wolframalpha was \[\sum_{n=0}^{\infty}\frac{n!}{3n+2}\] and I'm saying that that actually isn't equal to the given series. So like, a series is the sum of n terms of some sequence a(n) right? And I think that the above series is for a different a(n) given because it says each of the numbers on the bottom are multiplied not added.

Answer 20

And wolframalpha never loads for me :/ don't know why, it just doesn't. The link you posted is still loading lol. (or I suppose more appropriately failing to load)

Answer 21

Alright, I understand that, so what you did was computing the following values, and please check if I did any mistakes:
\[\Large a_1=\frac{1}{2} \\
\Large a_2=\frac{2}{8} \\
\Large a_3=\frac{6}{11} \\ \Large a_4=\frac{24}{14} \]

Answer 22

I believe so. because \[a_{1} = \frac{0!}{3n+2}=\frac{1}{2}\]\[a_{2}=\frac{1!}{(2)(3n+2)}=\frac{1}{10}\] and \[a_{3}=\frac{2!}{(2)(5)(3n+2)}=\frac{2}{80}\]

Answer 23

I suppose that should have started at a(0) not a(1)

Answer 24

Oh my bad, pardon me, I was all lost because of my attempt to sum the individual terms up, nevertheless, with the given information about the sum, which is our initial point (or at least mine from the opening post) I don't see how the individual terms of the sequence can be computed / deduced. Is it possible that your original problem is different from your opening post?

What I certainly do understand is that:

\[\Large S_n= \sum_{n=1}^N\frac{n!}{3n+2} \] is a sum to me, when there's an \(\infty\) as the upper level, I am used to immediately check from the given statement, to check if it converges or diverges. But if the above sum stands for the partial terms, then things look of course differently and I dramatically changes the entire topic (-:

Answer 25

The original question is \[\sum_{n=0}^{\infty}\frac{n!}{2*5*8*...*(3n+2)}\]

Answer 26

yes and the problem that emerges is about the multiplication of the denominator, I see.

Answer 27

So, we keep multiplying the previous bottom terms by the new term for each new term in the sequence. ...not sure how to phrase that better (i.e. not use term for three different things) but hopefully you understand what I'm saying.

(how do you make the equation appear larger? off-topic but curious)

Answer 28

yes

Answer 29

Use \Large for making it bigger you can also type it as \large, which is still large but not as big as \Large or \LARGE then there's also the same game with \huge \Huge \HUGE

Answer 30

ah. interesting.

Answer 31

good so we have:
\[\Large a_0=\frac{1}{2} \\
\Large a_1=\frac{1}{10} \\\Large a_2=\frac{2}{80} \\ \Large a_3=\frac{6}{880} \] And then you're quite right about a_n

Answer 32

So I also have: \[\Large a_n=\frac{na_{n-1}}{3n+2} \]
This definition doesn't clearly hold for \(n=0\), but I couldn't figure out a better way to define it myself. From there, you'd have to ask someone else, if the way you apply the ratio test is correct.

Answer 33

Okay. Awesome. Thanks. I think in that case I did, because early on when we first started sequences and series my book would sometimes define the first term of a sequence and then define the later terms like the one above. But they may have only started at 1 so not to sure there.

Answer 34

The best thing I could recommend you to do is repost this question and share your work in your opening post (more precisely, the post followed by it because the opening post never supports \(\LaTeX\))

I bet there a quite a few people that could help you from there, I clearly underestimated this problem at the beginning and got distracted by the sum rather by the sequence. However, your approach seems right to me.

Answer 35

Okay, I will probably do that tomorrow then since it's getting late here. Thanks for the advice!