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Math Help - Convergence & proof that e^x is greater than x^2 and such

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    Convergence & proof that e^x is greater than x^2 and such

    I have a few problems concerning infinite series which I need some help on. I'm supposed to determine the convergence/divergence using ONLY the comparison tests. Here are the problems:

    \displaystyle\Sigma\frac{1}{e^n+1}

    \displaystyle\Sigma\frac{2^n}{3^n-1000}

    \displaystyle\Sigma\frac{1}{e^n^2}

    For the first and the third ones, I want to apply the plain comparison test, using \frac{1}{x^2} as the series which I compare it to. However, I need a way to prove that e^n is greater than x^n for all n [1, \infty). Once I prove this, how would I apply it to the third problem since it's {e^{n}}^2? By the way, that is e^n^2...it's kind of hard to tell with the latex. For the second problem, here is what I have so far:

    \displaystyle\Sigma\frac{2^n}{3^n-1000}

    \displaystyle\frac{2^n}{3^n-1000}\geq\frac{2^n}{3^n}

    So, now using the limit comparison test...

    \displaystyle\lim_{n\to\infty}\frac{\frac{2^n}{3^n-1000}}{\frac{2^n}{3^n}} = \displaystyle\lim_{n\to\infty}\frac{3^n}{3^n-1000}=1

    The limit exits, is finite, and is greater than zero...so the original series will behave the same as \Sigma\frac{2^n}{3^n} but I'm not sure how to prove whether this series converges/diverges? Any help would be appreciated.
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    Quote Originally Posted by dbakeg00 View Post
    I have a few problems concerning infinite series which I need some help on. I'm supposed to determine the convergence/divergence using ONLY the comparison tests. Here are the problems:

    \displaystyle\Sigma\frac{1}{e^n+1}

    \displaystyle\Sigma\frac{2^n}{3^n-1000}

    \displaystyle\Sigma\frac{1}{e^n^2}
    [tex]e^{n^2} [/tex] gives  e^{n^2}

    Note that \dfrac{1}{e^n+1}<\dfrac{1}{2^n}
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    Quote Originally Posted by Plato View Post
    [tex]e^{n^2} [/tex] gives  e^{n^2}

    Note that \dfrac{1}{e^n+1}<\dfrac{1}{2^n}
    Thanks..I wasn't sure how to write that properly in Latex

    and \Sigma\frac{1}{2^n} converges because it s a geometric series...I should have seen that!

    Thank you

    Now can someone help me prove that e^x \geq x^2 for all x\geq 1?
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    Quote Originally Posted by dbakeg00 View Post
    Now can someone help me prove that e^x \geq x^2 for all x\geq 1?
    Let f(x)=e^x-x^2 show that is increasing for x>1.
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    Quote Originally Posted by dbakeg00 View Post
    Thanks..I wasn't sure how to write that properly in Latex

    and \Sigma\frac{1}{2^n} converges because it s a geometric series...I should have seen that!

    Thank you

    Now can someone help me prove that e^x \geq x^2 for all x\geq 1?
    \displaystyle \sum_{n \geq 0}{\frac{1}{2^n}} = \sum_{n \geq 0}{\left(\frac{1}{2}\right)^n} converges because it is a geometric series with \displaystyle |r| < 1.
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