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Math Help - Past ACTM question. Limits.

  1. #1
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    Past ACTM question. Limits.

    \lim_{x\rightarrow\infty}\left (\frac{1+2^{199}+3^{199}+4^{199}...+n^{199}}{n^{20  0}}\right )

    Is equal to:
    a. 1/201
    b.1/200
    c.1/199
    d.1/98
    e. None of these

    My attempt:
    I have no clue how to take this limit so I just tried to use calculator guessing.
    \frac{1}{1^{200}}=1

    \frac{1+2^{199}}{2^{200}}=.5

    \frac{1+2^{199}+3^{199}}{3^{200}}=.3333...

    \frac{1+2^{199}+3^{199}+4^{199}}{4^{200}}=.25...

    And my calculator wont go past this so I just figured that it continued with the pattern that it has now.

    so
    keeping with the pattern:
    if n=1
    it's
    1^{-1}
    if n=2
    it's
    2^{-1}
    if n=3
    it's
    3^{-1} ect.
    it must be zero because 1/infinity=0
    so e


    The answer however is B. Does anyone know how to work this limit?
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  2. #2
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    The method I used is recognizing the limit is actually a definite integral (in the form of a Riemann sum). (I wouldn't be surprised if someone else posted an easier solution involving a subtle algebraic manipulation of the actual limit instead of converting it to an integral).

    Maybe you remember that a definite integral is defined as a Riemann sum. If not, then here is the definition of a definite integral:

    \displaystyle \int_a^b f(x) \, dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x_i

    For the rest of this post, I am going to show that your problem is actually a Riemann sum (the right-hand side of the above equation). By doing so, I am allowed to convert it to an equivalent definite integral, which will hopefully be easy to evaluate!

    If the question were to find the area between f(x) = x^199 and the x-axis on the interval [a,b] using a Riemann sum with n rectangles, what would you do? The Riemann sum would look like:

    \displaystyle \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x_i.

    If we decide each of our rectangles will have the same width, then

    \Delta x_i = \Delta x = \frac{b-a}{n}.

    And then, x_i = a + \Delta x \cdot i = a + \frac{b-a}{n} i.

    Substituting these values into the Riemann sum gives:

    \displaystyle \lim_{n\to\infty} \sum_{i=1}^n \left [ \left ( a + \frac{b-a}{n} i \right ) ^{199} \cdot \frac{b-a}{n} \right ]

    This looks nothing like the original problem, right? Well, the first term of our summation is (i = 1):

    (a + \frac{b-a}{n})^{199} \cdot \frac{b - a}{n}.

    If we look back at the original problem, we can imagine the first term being \frac{1}{n^{200}}, which is equal to \left ( \frac{1}{n} \right ) ^{199} \cdot \frac{1}{n}. They are starting to look similar (a little!).

    Now, the natural question is what values of a and b will make the first term in the Riemann sum and the first term in the original problem match exactly?

    You may already notice that (a + \frac{b-a}{n})^{199} will probably have to match \left ( \frac{1}{n} \right ) ^{199}. Similarly, \frac{b-a}{n} will probably have to match \frac{1}{n}. Let's focus on (a + \frac{b-a}{n})^{199} and \left ( \frac{1}{n} \right ) ^{199}.

    \left ( \frac{1}{n} \right ) ^{199} doesn't have any addition. It isn't \left (10 + \frac{1}{n} \right ) ^{199}, it's just \left ( \frac{1}{n} \right ) ^{199}. If a = 0, then (a + \frac{b-a}{n})^{199} becomes \left ( \frac{1}{n} \right ) ^{199}. They match!

    Now, let's switch our focus to \frac{b-a}{n} and \frac{1}{n}. We now know a = 0, so:

    \frac{b}{n} will have to match \frac{1}{n}.

    We can conclude b = 1. Substituting a and b into the Riemann sum and checking other terms, such as i = 2 or i = 3, shows that our Riemann sum now matches the original problem exactly. In other words, I have just shown that your problem is actually a Riemann sum with a = 0 and b = 1. According to the definition of a definite integral, a Riemann sum is just a definite integral. Therefore, your problem is a definite integral!

    Now we know the original problem is equivalent to evaluating \int_a^b x^{199} \, dx, where a = 0 and b = 1.

    \displaystyle \lim_{n\to\infty} \dfrac{1+2^{199}+3^{199}+4^{199}...+n^{199}}{n^{20  0}} = \int_0^1 x^{199} \, dx = \frac{x^{200}}{200} \bigg |_0^1 = \frac{1}{200}

    Like I said, I wouldn't be surprised if someone now posts a much more elegant solution that has eluded me. Until then, this is the best I've got.

    In summary, the original problem is actually a definite integral. Once the limits of integration were found, it was only a matter of evaluating an integral.

    Hi, Jose. <3

    \displaystyle \int_a^b f(x) \, dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x_i
    Last edited by NOX Andrew; February 26th 2011 at 05:30 PM.
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