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Math Help - Convergence of a Series Help

  1. #1
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    Convergence of a Series Help

    I'm asked to determine whether the following series converges or diverges: \displaystyle{\sum_{n=0}^{\infty}\frac{n!}{e^{n^{2  }}}}

    My solution:

    \displaystyle{\sum_{n=0}^{\infty}\frac{n!}{e^{n^{2  }}} = {\sum_{n=0}^{\infty}\frac{n!}{e^{2n}}.

    Therefore we have by the ratio test for convergence (and ignoring absolute value since all terms are already positive):

    lim_{n\rightarrow\infty}\frac{(n+1)!\cdot e^{2n}}{e^{(n+1)^2}\cdot n!}= lim_{n\rightarrow\infty}\frac{(n+1)\cdot n! \cdot e^{2n}}{e^{2n}\cdot e^2 \cdot n!} = lim_{n\rightarrow\infty}\frac{n+1}{e^2} = \infty

    Therefore the series diverges by the ratio test.

    However, it apparently converges according to the answer in the back of my text (I guess e^{n^2} end up growing much faster than n! which is kind of surprising to me because obviously n! grows much faster than just e^n for even relatively small n). In any case, I'd appreciate it if someone can show me the correct solution.

    In calculus II the ratio test is the only test we have for convergence of a series involving factorials. I suspect my problem was purely algebraic.

    Thanks ~

    By the way, when people write e^{(n+1)^2} do they mean distribute the 2 across the originally raised expression (as would normally be the case following the law of exponents) or do they mean to actually square the originally raised expression?

    In other words, is the above expression equivalent to

    1.) e^{(n+1)^2} = e^{2n+2} = e^{2n}\cdot e^2
    2.) e^{(n+1)^2} = e^{n^2+2n+1} = e^{n^2}\cdot e^{2n}\cdot e^1 = e^{2n}\cdot e^{2n}\cdot e = e^{4n} \cdot e

    I'm pretty sure it has to be the first.....
    Last edited by TaylorM0192; July 19th 2010 at 04:19 PM.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    First of all.. in your third line, why it equals?!

    Second, I think this is will be helpful: Stirling's approximation - Wikipedia, the free encyclopedia
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  3. #3
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    Oh, I was just simplifying the exponent. I feel like I might have a complete misunderstanding with basic algebra here....

    e^{n^2} = e^{2n} correct?

    And thank you for the post, but I don't think it's exactly what I am looking for. I'm not so much interested in computing factorials for large n - more so in comparing the respective magnitudes of the factorial and exponential function (so I can see why it actually converges according to the 'correct answer') when the exponential function's argument is squared (obviously the factorial definitely grows much faster than the oridinary exponential function).

    Thanks though~
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  4. #4
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    Quote Originally Posted by TaylorM0192 View Post
    I'm asked to determine whether the following series converges or diverges: \displaystyle{\sum_{n=0}^{\infty}\frac{n!}{e^{n^{2  }}}}

    My solution:

    \displaystyle{\sum_{n=0}^{\infty}\frac{n!}{e^{n^{2  }}} = {\sum_{n=0}^{\infty}\frac{n!}{e^{2n}}.

    Therefore we have by the ratio test for convergence (and ignoring absolute value since all terms are already positive):

    lim_{n\rightarrow\infty}\frac{(n+1)!\cdot e^{2n}}{e^{(n+1)^2}\cdot n!}= lim_{n\rightarrow\infty}\frac{(n+1)\cdot n! \cdot e^{2n}}{e^{2n}\cdot e^2 \cdot n!} = lim_{n\rightarrow\infty}\frac{n+1}{e^2} = \infty

    Therefore the series diverges by the ratio test.
    Your formula has e^{n^2} but then you have e^{2n}. Those are not the same. (e^n)^2= e^{2n} but not e^{n^2}. As an obvious example, if n= 3, e^{3^2}= e^9 which is about 8103 while e^{2(3)}= e^6 which is about 403. Note that \left(e^3\right)^2= (20.08)^2 is about 403.

    What you want is \frac{(n+1!}{e^{(n+1)^2}}\frac{e^{n^2}}{n!} = \frac{(n+1)!}{n!}\frac{e^{n^2}{e^{(n+1)^2}} = \frac{n+1}{1}\frac{e^{n^2}}{e^{n^2+ 2n+ 1}}= \frac{n+1}{1}\frac{e^{n^2}}{e^{n^2}(e^{2n})(e)}= \frac{n+1}{e(e^n)}.

    However, it apparently converges according to the answer in the back of my text (I guess e^{n^2} end up growing much faster than n! which is kind of surprising to me because obviously n! grows much faster than just e^n for even relatively small n). In any case, I'd appreciate it if someone can show me the correct solution.

    In calculus II the ratio test is the only test we have for convergence of a series involving factorials. I suspect my problem was purely algebraic.

    Thanks ~

    By the way, when people write e^{(n+1)^2} do they mean distribute the 2 across the originally raised expression (as would normally be the case following the law of exponents) or do they mean to actually square the originally raised expression?

    In other words, is the above expression equivalent to

    1.) e^{(n+1)^2} = e^{2n+2} = e^2n\cdot e^2
    1.) e^{(n+1)^2} = e^{n^2+2n+1} = e^{n^2}\cdot e^{2n}\cdot e^1 = e^2n\cdot e^2n\cdot e = e^{4n+1}

    I'm pretty sure it has to be the first.....
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  5. #5
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    Thanks! I suspected as much that it was a problem in misunderstanding what was meant by the expression e^{n^2}

    Makes sense now.
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