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Test series

  1. May 8, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    Test the series for convergence or divergence[tex]\sum_{n=1}^{\infty} \frac{n!}{4.7.10...(3n+1)}x^n,\; x>0[/tex]

    The attempt at a solution

    I've decided to use the ratio test because of the factorial.
    [tex]L=\lim_{n\to \infty}\frac{u_{n+1}}{u_n}[/tex]
    I worked it out and i got:
    [tex]L=\lim_{n\to \infty}\frac{4x(3n+1)(n+1)}{3n+4}=\lim_{n\to \infty}\frac{4x(3n^2+4n+1)}{3n+4}[/tex] Then, i did long division for the n terms.
    [tex]\lim_{n\to \infty} 4x \left( n+ \frac{1}{3n+4}\right)=4x\lim_{n\to \infty} \left( n+ \frac{1}{3n+4}\right)=4x(∞+1/∞)=∞[/tex]
    According to the ratio test, since L>1, the series diverges.

    Is this correct?
     
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  3. May 8, 2012 #2

    LCKurtz

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    You ratio test should have an |x| in there. But something else is wrong with your ratio. Where did the 4 and the (3n+1) come from?
     
  4. May 8, 2012 #3

    HallsofIvy

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    Perhaps if you wrote the multiplication of the fractions out in detail, it would be clearer.
     
  5. May 8, 2012 #4

    Mark44

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    Show us how you arrived at this result. As LCKurtz pointed out, there should be a factor of |x| in it.
     
  6. May 9, 2012 #5

    sharks

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    [tex]u_n=\frac{n!x^n}{4.7.10...(3n+1)}[/tex]
    [tex]u_{n+1}=\frac{(n+1)!x^{n+1}}{4.7.10...(3(n+1)+1)}[/tex]
    [tex]\frac{u_{n+1}}{u_n}=\frac{(n+1)!x^{n+1}}{4.7.10...(3(n+1)+1)}\times \frac{4.7.10...(3n+1)}{n!x^n}[/tex]
    [tex]\frac{u_{n+1}}{u_n}=\frac{(n+1)n!x^nx}{4.7.10...(3(n+1)+1)}\times \frac{4.7.10...(3n+1)}{n!x^n}[/tex]
    OK, i realize that i had made a mistake previously, but i'm still not sure:[tex]\frac{u_{n+1}}{u_n}=\frac{(n+1)x}{(3(n+1)+1)}= \frac {(n+1)x} {(3n+4)}[/tex]Doing long division:[tex]\frac {(n+1)x} {(3n+4)}=\frac{1}{3}x-\frac{x}{3(3n+4))}[/tex]Taking the limit, gives:[tex]\lim_{n\to\infty} \left[ \frac{1}{3}x-\frac{x}{3(3n+4)}\right]= \frac{1}{3}x[/tex]
     
    Last edited: May 9, 2012
  7. May 9, 2012 #6

    HallsofIvy

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    Yes, that is correct- although again, you have dropped the absolute value sign on x.

    Now, do you remember why you divided those fractions? You wanted to use the ratio test. What does the ratio test say?
     
  8. May 9, 2012 #7

    Mark44

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    <snip>
    Long division is unnecessary.

    $$|x|\frac{n+1}{3n + 4} = |x|\frac{n(1 + 1/n)}{n(3 + 4/n)}$$
     
  9. May 9, 2012 #8

    sharks

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    I have a few doubts/questions:

    1) I conclude that the requirement of having the absolute value sign on x is due to [itex]x > 0[/itex] in the original problem. In that case, the absolute value sign should appear as from the first line in post #5, as [itex]|x^n|[/itex], and in the second line, as [itex]|x^{n+1}|[/itex]. Correct?

    2) Instead of [itex]\frac{1}{3}|x|[/itex], could i have alternatively given the answer as [itex]\frac{1}{3}x,\; x>0[/itex]? This should mean the same thing.

    3) Now that i know the evaluation of the limit is [itex]\frac{1}{3}|x|[/itex], i need to find the value of L to know if the series converges or diverges.
    However, here is the problem: I do not know the value of x. If x>3, then L>1 and the series diverges, but here is another possibility: if x=3, then L=1 and this means the series can either converge or diverge. A third possibility is if x<3, then L<1 and the series converges.
    So, my question is: how do i know which value will x have? Although, x>0 does seem to indicate that i can take it to be as large a value as i like, in which case, L>1.

    After some more thinking, i figure that as x increases from 0 to a value greater than 3, the series first converges, then diverges. But i'm not sure what the final answer should be.
     
    Last edited: May 9, 2012
  10. May 9, 2012 #9

    Mark44

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    There's nothing in the original problem that says x > 0. In the Ratio Test, you look at this limit:
    $$\lim_{n \to \infty}\left| \frac{a_{n+1}}{a_n}\right|$$

    Yes, but you can write these as |x|n and |x|n+1.
    Bad idea since you are potentially losing half the interval of convergence.
    If |x|/3 < 1 the series converges absolutely. What interval does that represent?
    If |x|/3 = 1, the Ratio Test is inconclusive. Substitute the two values of x into the original series, and use some other test to determine the convergence/divergence of each of these two endpoints.
     
  11. May 9, 2012 #10

    sharks

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    Since x cannot be a negative value and the original problem states that x>0, [tex]0<|x|<3[/tex]
    The two values of x must be: <3 and >3. I choose x=1 and x=4. I'm not sure what you mean here.
     
    Last edited: May 9, 2012
  12. May 9, 2012 #11

    Mark44

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    Apologies - I missed that it says x > 0 in your OP.

    |x|/3 < 1 ==> -3 < x < 3
    Since we also have x > 0, the interval reduces to 0 < x < 3, so I will dispense with absolute values.

    If x/3 > 1, the Ratio Test says the series diverges (i.e., if x > 3).

    If x/3 = 1, or equivalently, if x = 3, the Ratio Test is inconclusive. Use some other test to determine whether the series converges or diverges at this value.
     
  13. May 9, 2012 #12

    sharks

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    So, my final answer will have to comprise of 3 parts: when 0<x<3, x>3 and x=3. Correct?

    Now, at x=3, the series becomes: [tex]\sum_{n=1}^{\infty} \frac{n!}{4.7.10...(3n+1)}3^n[/tex]I would usually immediately go for the Ratio Test, but since i have to use another test... I would say the comparison test or the limit comparison test. But i don't know with what expression to compare it to, due to the n! term. I think the nth-term test might be easier, but then i would need to differentiate the numerator which contains n! and what's the derivative of n! w.r.t.n?
     
  14. May 9, 2012 #13

    Mark44

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    If you're thinking "derivative" you're probably thinking L'Hopital's Rule, which is the wrong direction to go, IMO.

    Break up the general term of the series like this...
    $$\frac{1}{4} \frac{2}{7} \frac{3}{10}... \frac{n}{3n + 1}\cdot~3^n $$

    Each of the fractions is bounded above and below, and there are n of them. Write an inequality with 3 members, with the above in the middle, and then take the limit of all three members.
     
  15. May 9, 2012 #14

    sharks

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    In your expression above, the numerator should be n! not just n, right?

    The sequence of partial sums:[tex]S_n=\frac{3}{4}+\frac{18}{28}+\frac{162}{280}+...[/tex]
    The sequence:[tex]a_n=\frac{n!}{3n + 1}\cdot~3^n=\frac{3}{4},\frac{18}{7},\frac{54}{10},...[/tex]
    I have no idea what i'm doing.
     
    Last edited: May 9, 2012
  16. May 9, 2012 #15

    Mark44

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    No. I have broken n! up into 1*2*3* ... *n.
     
  17. May 9, 2012 #16

    sharks

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    OK, just to clarify your suggestion, for n=3:
    [tex]\frac{1}{4} \frac{2}{7} \frac{3}{10}... \frac{n}{3n + 1}\cdot~3^n=\frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}... \frac{n}{3n + 1}\cdot~3^n[/tex] But in your expression, [itex]3^n[/itex] is not multiplied into the product. It should have been:[tex]\left( \frac{1}{4}\times \frac{2}{7}\times \frac{3}{10}\right)\cdot~3^3... \frac{n}{3n + 1}\cdot~3^n[/tex]
     
    Last edited: May 9, 2012
  18. May 9, 2012 #17

    Mark44

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    Yes, those are the same. I debated with myself about putting parentheses around each fraction, but decided it wasn't necessary.
    No it shouldn't. That's not the way multiplication works. You are apparently thinking that there is some distributive product for multiplication. There isn't.

    IOW, this is NOT TRUE: (ab)c = (ac)(bc)

    Multiplication distributes over addition ( a(b + c) = ab + ac) , but it doesn't distribute over multiplication!
     
  19. May 9, 2012 #18

    sharks

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    But i don't understand how to write an inequality out of these. Here's an attempt:
    [tex]\left( \frac{1}{4}<\frac{2}{7}<\frac{3}{10}<\frac{n}{3n + 1}\right)3^n[/tex]
     
    Last edited: May 9, 2012
  20. May 9, 2012 #19

    Mark44

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    For a start, let's focus on the fractions:
    (1/4)(2/7)(3/10)...(n/(3n + 1))

    1. How many fractions are in the above?
    2. Every fraction above is >= what number?
    3. Hence, the product above is >= what number?
    4. Every fraction above is < what number?
    5. Hence, the product above is < what number?

    Now can you write an inequality involving (1/4)(2/7)(3/10)...(n/(3n + 1))3n?
     
  21. May 9, 2012 #20

    sharks

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    1. There are 4 fractions.

    2. [itex]\frac{2}{7}>\frac{1}{4},\; \frac{3}{10}>\frac{2}{7},\; \frac{n}{3n + 1}>\frac{3}{10}[/itex]

    3. Hence, [itex]\frac{n}{3n + 1}>\frac{1}{4}[/itex]

    4. Every fraction above is less than [itex]3^n[/itex]

    5. Hence, [itex]\frac{n}{3n + 1}< 3^n[/itex]

    The inequality is:
    [tex]\frac{1}{4}< \frac{n}{3n + 1}<3^n[/tex]
    If the above inequality is correct, then taking the limits for all 3 members:
    [tex]\lim_{n \to \infty}\frac{1}{4}=\frac{1}{4}[/tex]
    [tex]\lim_{n \to \infty}\frac{n}{3n + 1}=\lim_{n \to \infty}\frac{1}{3 + 1/n}=\frac{1}{3}[/tex]
    [tex]\lim_{n \to \infty}3^n=∞[/tex]The first two limits converges and the third limit diverges. What does this mean? The squeeze theorem fails in this case.
     
    Last edited: May 9, 2012
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