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Test series for convergence or divergence

  1. May 8, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    There are 3 parts to this problem:
    [tex](a) \; \sum^{\infty}_{n=1} \frac{n^4}{4^n}[/tex]
    [tex](b) \; \sum^{\infty}_{n=1} \left( \frac{n+8}{n} \right)^n[/tex]
    [tex](c) \; \sum^{\infty}_{n=1} \frac{5^n-8}{4^n+11}[/tex]

    The attempt at a solution

    (a) I've used the Ratio test.

    So, [itex]u_n=\frac{n^4}{4^n}[/itex] and [itex]u_{n+1}=\frac{(n+1)^4}{4^{(n+1)}}[/itex]
    [tex]L=\lim_{n\to \infty} \frac{u_{n+1}}{u_n}=\frac{1}{4}\lim_{n\to \infty} \left( \frac{n+1}{n} \right)^4[/tex]
    I don't know what to do at this point.

    (b) It looks just like the last step that i reached with the above. I'm thinking that it resembles a geometric series with [itex]r=\frac{n+8}{n}[/itex] but i can't evaluate r to show that it lies between -1 and 1.

    (c) I used the comparison test. It failed, so i used the limit comparison test and got stuck.

    For n→∞, the limit becomes [tex]\frac{5^n-8}{4^n}[/tex]
     
    Last edited: May 8, 2012
  2. jcsd
  3. May 8, 2012 #2

    Mark44

    Staff: Mentor

    $$=\frac{1}{4}\lim_{n\to \infty} \left( \frac{n^4(1+1/n)^4}{n^4} \right)$$
    Can you continue from here?
    Show us what you got.
    No, it doesn't. When you take the limit, anything involving n will get evaluated. After taking the limit, n should not appear.

    This problem (the third part a) has an obvious answer. Maybe not obvious to you right now, but when you see it, you'll want to slap your forehead.:tongue:
     
    Last edited: May 8, 2012
  4. May 8, 2012 #3

    sharks

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    But your suggestion [itex]\left( \frac{n^4(1+1/n)^4}{n^4} \right)[/itex] is not equivalent to [itex]\left( \frac{n+1}{n} \right)^4[/itex]
    Since, [itex]\left( \frac{n+1}{n} \right)^4= \frac{(n+1)^4}{n^4}[/itex]

    (b) [itex]\sum^{\infty}_{n=1} \left( \frac{n+8}{n} \right)^n=\sum^{\infty}_{n=1} \frac{(n+8)^n}{n^n} [/itex] and then i don't know, especially since [itex]n^n[/itex] is very intimidating.

    (c) I meant, for n large, the comparison can be done with: [itex]\frac{5^n-8}{4^n}[/itex]
     
    Last edited by a moderator: May 8, 2012
  5. May 8, 2012 #4

    Mark44

    Staff: Mentor

    I omitted an exponent of 4 in my previous post, but it should be fixed now.
    $$\left( \frac{n+1}{n} \right)^4 = \frac{(n+1)^4}{n^4} $$

    What I did was factor n out of each term in the numerator on the right. When you bring n out of the parentheses, it becomes n4.
    No sense in leaving the "-8" in the numerator. For large n, ##\frac{5^n-8}{4^n + 11} \approx \frac{5^n}{4^n}##
     
  6. May 8, 2012 #5

    Mark44

    Staff: Mentor

    One thing you should be doing, but probably aren't, is looking at the behavior of the general term of a series as n gets large.

    If ##\lim_{n \to \infty}a_n = 0##, then you can't tell anything about the series.

    But, if ##\lim_{n \to \infty}a_n \neq 0##, then you can definitely say something about the series.
     
  7. May 8, 2012 #6

    sharks

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    OK, for part (a), the answer is: 1/4. Then, since L<1, the series converges.

    Part (b):
    [tex]\sum^{\infty}_{n=1} \frac{(n+8)^n}{n^n}=\sum^{\infty}_{n=1} \frac{n^n(1+8/n)^n}{n^n}=\sum^{\infty}_{n=1} (1+8/n)^n[/tex] and i'm stuck again.

    Part (c): Let [itex]u^n=\sum^{\infty}_{n=1} \frac{5^n-8}{4^n+11}[/itex]

    Then [itex]v^n=\sum^{\infty}_{n=1} \frac{5^n}{4^n}=\sum^{\infty}_{n=1} (\frac{5}{4})^n[/itex] and it diverges as it is a geometric series with r>1

    Since, [itex]u_n ≤ v_n[/itex] and [itex]v_n[/itex] diverges, the ratio test fails. What to try next?

    Somehow, i can't seem to fully grasp series. The solution almost always seems elusive. I guess i just have to practice more.
     
  8. May 8, 2012 #7

    Mark44

    Staff: Mentor

    Yes.
    Take the limit of the n-th term. This is easier said than done, as it involves logs and L'Hopital's Rule.
    Take the limit of the n-th term. This limit is way easier than the one in part b.
     
  9. May 8, 2012 #8

    sharks

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    For part (b), what if i proceeded like this:
    [itex]\sum^{\infty}_{n=1} (1+8/n)^n=(1+8/∞)^∞[/itex] but then i evaluate whatever is inside the brackets first, so, [itex](1+8/∞)^∞=(1+0)^∞=(1)^∞=1[/itex]. Is this correct?
    Ultimately, i could simply just use the nth-term test and the series can be immediately confirmed as divergent.

    For part (c), applying the nth-term test, immediately, it becomes obvious that [itex]\sum^{\infty}_{n=1} \frac{5^n-8}{4^n+11}= \frac{5^∞-8}{4^∞+11}=∞[/itex] so, the series is divergent.

    However, this seems too easy and i don't really know when the nth-term test is applicable, as it seems to be able to work its way into anything.
     
    Last edited: May 8, 2012
  10. May 8, 2012 #9

    Mark44

    Staff: Mentor

    Substituting ∞ is not allowed.
    No, your limit value is incorrect.

    To evaluate lim (1 + 8/n)^n, let y = (1 + 8/n)^n, then take the limit. You'll need to use L'Hopital's Rule, as I mentioned earlier.
    This is premature, since you haven't correctly evaluated the limit yet.
    Again, you can never just substitute ∞ in an expression.

    Also,
    $$\sum^{\infty}_{n=1} \frac{5^n-8}{4^n+11} \neq \frac{5^n-8}{4^n+11} $$

    You are saying that the entire sum equals one of its terms. Not true.

    The way you did it, yes.
    The n-th term test is applicable to any series. The kicker is that you often can't conclude anything from it.
     
  11. May 8, 2012 #10

    sharks

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    OK, first i'll work on part (c):
    From [itex]\sum^{\infty}_{n=1} \frac{5^n-8}{4^n+11}[/itex], i'll take the limit.

    So, [itex]\lim_{n\to \infty} \frac{5^n-8}{4^n+11}[/itex] requires use of the l'Hopital's rule, giving: [itex]\lim_{n\to \infty} \frac{5^n.\ln 5}{4^n.\ln 4}[/itex]

    Now, if i group the logs together: ln (5-4) gives ln 1 which is equal to 0. Hence, the limit is 0.
    By the nth-term test, we cannot say if the series converges or diverges. This looks like another dead end.
     
  12. May 8, 2012 #11

    Mark44

    Staff: Mentor

    L'Hopital's Rule is not required. Factor 5n out of the numerator, and factor 4n out of the denominator.
    No. ln(5)/ln(4) ≠ ln(5 - 4). It is true that ln(5/4) = ln(5) - ln(4) though.
    Don't you have any sense of what the terms in the series are doing?
     
  13. May 8, 2012 #12

    sharks

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    Ah yes, i messed up my logs. Sorry about that.

    So, [itex]\lim_{n\to \infty} \frac{5^n.\ln 5}{4^n.\ln 4}=\lim_{n\to \infty} \frac{\ln 5}{\ln 4}(\frac{5}{4})^n[/itex]

    Here, [itex]r=\frac{5}{4}[/itex] and since r>1, the limit diverges (not equal to zero). Therefore, (finally!) according to the nth-term test, the series diverges.

    For part (b), previously i had: [tex]\sum^{\infty}_{n=1} \frac{(n+8)^n}{n^n}=\sum^{\infty}_{n=1} \frac{n^n(1+8/n)^n}{n^n}=\sum^{\infty}_{n=1} (1+8/n)^n[/tex]
    But that method leads to nowhere, if i'm to use the nth-term test.
    So, going back and reworking my way, taking the limit and using l'Hopital's rule: [tex]\lim_{n\to \infty} \frac{(n+8)^n}{n^n}=
    \frac{\frac{n(n+8)^n}{(n+8)}+\ln (n+8).(n+8)^n}{n^n( \ln(n) + 1)}[/tex]Up to this point, is it correct?
     
    Last edited: May 8, 2012
  14. May 8, 2012 #13

    Mark44

    Staff: Mentor

    Yes! Good!
    No, it does lead somewhere.

    Go back a few posts (post #9) and see what I said about how to carry this out.
    That looks awful! It's much simpler to work with (1 + 8/n)n.
     
  15. May 8, 2012 #14

    sharks

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    [tex]Let\; y = \left(1 + \frac{8}{n}\right)^n[/tex]
    Then, taking logs on both sides:
    [tex]\ln y = \ln \left(1 + \frac{8}{n}\right)^n=n.\ln \left(1 + \frac{8}{n}\right)=
    \frac{\ln \left(1 + \frac{8}{n}\right)}{1/n}[/tex]
    Now, applying l'Hopital's rule gives:
    [tex]\ln y =\frac{8}{1+8/n}[/tex]
    Now, i have to get y on the L.H.S.:
    [tex]y =e^{\frac{8}{1+8/n}}[/tex]
    Thus,
    [tex]\lim_{n\to \infty} y =e^{\frac{8}{1+8/n}}=e^8[/tex]
    The limit exists but is not equal to zero, hence by the nth-term test, the series diverges. I hope my work above is correct. I'm not sure about the exponential part, but it has to be that way, since i need to express y in terms of n, right?
     
  16. May 8, 2012 #15

    Mark44

    Staff: Mentor

    Very nice!
    What you're really saying is this:
    [tex]\lim_{n \to \infty} \ln y =\lim_{n \to \infty}\frac{8}{1+8/n}[/tex]

    You can interchange the lim and ln operations, as long as the thing you're taking the log of is continuous at the point in question (ln(x) is defined for x as large as you'd like).

    This gives us
    [tex]\ln(\lim_{n \to \infty} y) =\lim_{n \to \infty}\frac{8}{1+8/n}[/tex]

    Eventually you get to ln (lim y) = 8, so lim y = e8.

     
  17. May 8, 2012 #16

    sharks

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    Thank you very much for your help and patience, Mark44. :smile:
     
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