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Math Help - Finding smallest n for finite sum

  1. #1
    Senior Member Pinkk's Avatar
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    Finding smallest n for finite sum

    Find smallest n so that the finite sum \sum_{k=1}^{n}\frac{(-2)^{k}}{k!} is certain to differ from \sum_{k=1}^{\infty}\frac{(-2)^{k}}{k!} by less than .001.

    So I set up the inequality:

    a_{n+1} \le .001

    \frac{(-2)^{n+1}}{(n+1)n!} \le \frac{1}{100}

    (n+1)n! \ge -200(-2)^{n}

    And that is where I am stuck. How do I solve for n?
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  2. #2
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    Quote Originally Posted by Pinkk View Post
    Find smallest n so that the finite sum \sum_{k=1}^{n}\frac{(-2)^{k}}{k!} is certain to differ from \sum_{k=1}^{\infty}\frac{(-2)^{k}}{k!} by less than .001.

    So I set up the inequality:

    a_{n+1} \le .001

    \frac{(-2)^{n+1}}{(n+1)n!} \le \frac{1}{100}

    (n+1)n! \ge -200(-2)^{n}

    And that is where I am stuck. How do I solve for n?
    note that a_{n+1}=\frac{2^{n+1}}{(n+1)!} not \frac{(-2)^{n+1}}{(n+1)!}. the inequality \frac{2^{n+1}}{(n+1)!}< .001 cannot be solved for n algebraically. however, it is possible to make it a little simpler by finding a lower bound for \frac{2^{n+1}}{(n+1)!},

    which is not worth the effort anyway. the easiest way, and you're expected to do it this way, to use your calculator. i think you'll get n=9. (don't trust my calculations! do it yourself! )
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  3. #3
    Senior Member Pinkk's Avatar
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    We're not allowed to use calculators for any of our exams, so my professor must have accidentally entered this problem. And why is it \frac{2^{n+1}}{(n+1)!} and not \frac{(-2)^{n+1}}{(n+1)!}?
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  4. #4
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    Quote Originally Posted by Pinkk View Post

    We're not allowed to use calculators for any of our exams, so my professor must have accidentally entered this problem.
    you won't have such a problem in your exam then! haha


    And why is it \frac{2^{n+1}}{(n+1)!} and not \frac{(-2)^{n+1}}{(n+1)!}?
    because a_n is always the absolute value of the n-th term of your alternating series. the theorem that they taught you, and you should take a look at it again, is that if the sequence \{a_k \} is

    positive, decreasing and \lim_{k\to\infty} a_k=0, then for any n: \ \left|\sum_{k=1}^{\infty} (-1)^k a_k - \sum_{k=1}^n (-1)^k a_k \right| \leq a_{n+1}.
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  5. #5
    Senior Member Pinkk's Avatar
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    Ah, okay, I get it. Thanks!
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