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Thread: Past ACTM question. Limits.

  1. #1
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    Past ACTM question. Limits.

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

    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.
    $\displaystyle \frac{1}{1^{200}}=1$

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

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

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

    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:

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

    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:

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

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

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

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

    Substituting these values into the Riemann sum gives:

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

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

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

    If we look back at the original problem, we can imagine the first term being $\displaystyle \frac{1}{n^{200}}$, which is equal to $\displaystyle \left ( \frac{1}{n} \right ) ^{199} \cdot \frac{1}{n}$. 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 $\displaystyle (a + \frac{b-a}{n})^{199}$ will probably have to match $\displaystyle \left ( \frac{1}{n} \right ) ^{199}$. Similarly, $\displaystyle \frac{b-a}{n}$ will probably have to match $\displaystyle \frac{1}{n}$. Let's focus on $\displaystyle (a + \frac{b-a}{n})^{199}$ and $\displaystyle \left ( \frac{1}{n} \right ) ^{199}$.

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

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

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

    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 $\displaystyle \int_a^b x^{199} \, dx$, where a = 0 and b = 1.

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

    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

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