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Thread: 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:

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

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

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

    For the first and the third ones, I want to apply the plain comparison test, using $\displaystyle \frac{1}{x^2}$ as the series which I compare it to. However, I need a way to prove that $\displaystyle e^n$ is greater than $\displaystyle x^n$ for all n [1,$\displaystyle \infty)$. Once I prove this, how would I apply it to the third problem since it's $\displaystyle {e^{n}}^2$? 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:

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

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

    So, now using the limit comparison test...

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

    The limit exits, is finite, and is greater than zero...so the original series will behave the same as $\displaystyle \Sigma\frac{2^n}{3^n}$ 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:

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

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

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

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

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

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

    Thank you

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

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

    Thank you

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