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Limits of Sequences

  1. Jul 14, 2007 #1
    I have already completed this calculus course but I cant seem to do these problems that I should know :grumpy: I have to find the limit of the sequence which seems to be the same as the limit of a function.

    1. The problem statement, all variables and given/known data

    Find the limit of the given sequence as [tex]n \rightarrow \infty [/tex]

    [tex] \frac{\ (-1)^n\sqrt{n+1}}{n}[/tex]

    Here the [tex] (-1)^n [/tex] is throwing me... Should I just take the absolute value or what?

    [tex]\frac{10^n}{n!}[/tex]

    Here my problem is the factorial. I want to use L'Hospital rule, but the factorial isnt differentiable. Im am only a little familiar with the gamma function, what should I do? I have a few with factorials that Im stuck on.




    Thanks
     
  2. jcsd
  3. Jul 14, 2007 #2
    Notice that,
    [tex]\left| \frac{(-1)^n \sqrt{n+1}}{n} \right| \leq \frac{\sqrt{n+n}}{n} = \frac{\sqrt{2}}{\sqrt{n}} \to 0[/tex]

    Use the Ratio Test:
    [tex] \lim_{n\to \infty} \frac{10^{n+1}}{(n+1)!}\cdot \frac{n!}{10^n} = \lim_{n\to \infty} \frac{10}{n+1}\to 0[/tex]
    Since the limit [tex]|L|<1[/tex] it means the original sequence converges to 0.
     
  4. Jul 14, 2007 #3
    OK, Thank you. Each of those make sense to me. I still wonder if there is another way however... Because the sections in the book that describe your two methods are after these problems.

    EDIT- BTW, the book is Math Methods, Boas, 3rd edition.
     
  5. Jul 14, 2007 #4

    Gib Z

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    Well those methods, Such as the Ratio test, are the ways of formally proving the fact, in the sections of the book earlier than them it may just have expected you to work out that the factorial function gets large much quicker than 10^n.

    And for the first one, alternatively you may remember that as x--> infinity, you only need to consider the leading term in a polynomial, as that is the dominating one. So as n-->infinity, n+1 ---> n.

    So ignoring the (-1)^n for now, it becomes [tex] \frac{ \sqrt{n} }{n} = \frac{1}{\sqrt{n}}[/tex], which as n --> infinity becomes 0. Knowing that all (-1)^n can do is change the sign, and that the sign doesn't matter for 0, we are done.
     
  6. Jul 15, 2007 #5

    HallsofIvy

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    Notice that you cannot, in general, just ignore (-1)n. n/(n+1) converges to 1 but (-1)nn/(n+1) does not converge. Do you see why?
     
  7. Jul 15, 2007 #6
    Okay

    Well, forget L'Hopital's Rule; hell, forget math for a second. With series, the math is intrinsically vital; with sequences, just visualizing what the hell is going on is a lot better.


    Okay, here goes:


    Visualize a square root function, and a linear function (the numerator and the denominator of the sequence). The square root function grows slowly; it takes its good old time to reach the same numbers that the linear function climbs to early on. So as you increase the value of x (or n), the denominator is increasing a lot faster than the numerator is, right? I mean, even though the numerator is shifted across the x-axis by one unit, that kinda goes away, doesn't it, as your numbers get larger?

    Take, for example, if you add 1 to 4 and then take the square root; that equals 2.23... and the other is 2. That seems like a good difference early on, right? Well, look at it as you increase the value of x: 100000 and 100001

    sqrt(1000000) = 316.2277
    sqrt(1000001) = 316.2293

    See how small that difference is getting? Well, taking x to infinity is like taking that difference and making it 0. It's related to the whole concept of if x = infinity, then "anything plus infinity is still infinity." So even though that x value on the top has one added to it, you can forget about it when we push this sequence to the extremes of infinity.

    So then, what happens when top and the bottom when we make that +1 negligable? Well, the bottom is growing faster. What happens when your denominator is growing faster than your numerator? Well, it gets smaller. What happens when you amplify that to the extremes of infinity? That seeming small increase to the denominator becomes infinite, and the whole function approaches 0.

    Well, that takes care of the whole oscillation problem, now doesn't it? You can't attach a sign to 0, can you? It's neutral, so whatever (-1)^infinity is, it doesn't matter. You can't calculate it's effect anyways.


    And that's how you've got to look at sequences. Let's apply this concept to your next sequence.

    10^n is growing really, really fast, right? But is n! growing faster?

    Well, look at it this way: 10^n is growing really fast, but n! is growing even faster. It's growing hyper fast, because it keeps on multiplying everything before it. 10*10*10 is fast, but keep in mind, n! is multiplying everything, including all of those values of 10*10*10, plus every number inbetween together. Much faster, no?

    So since the denominator is growing faster than the numerator, it approaches zero.
     
  8. Jul 15, 2007 #7

    Gib Z

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    yup that is true :) I justify myself that I didn't completely ignore them, but just noticed that they wouldn't make a difference if the rest of the expression goes to 0.
     
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