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Thread: Convergency of a series

  1. #1
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    Convergency of a series

    It is supposed to converge by the comparison test... Just anyone is able to explain me how? I'm getting crazy with this particular one...


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  2. #2
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    Re: Convergency of a series

    What exactly is the summation in the image?
    I see an m factorial but what is the rest of it?
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    Re: Convergency of a series

    Quote Originally Posted by Plato View Post
    What exactly is the summation in the image?
    I see an m factorial but what is the rest of it?
    What do u mean? The exercise wants me to prove that this series converges... (It's n factorial. There is no m)
    sum n!*ln^n(1+2/n)

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    Re: Convergency of a series

    Quote Originally Posted by domy7997 View Post
    What do u mean? The exercise wants me to prove that this series converges... (It's n factorial. There is no m)
    sum n!*ln^n(1+2/n) Sent from my Redmi Note 4X using Tapatalk
    to domy7997, you have very poor penmanship. Those are clearly m's in the image you posted.
    It may serve better if you used some devise other than a mobile phone.
    Also be reminded that this is an English language website.

    Now just to be sure that the series is: $\displaystyle \sum\limits_{n = 1}^\infty {n!{{\left[ {\log \left( {1 + \frac{2}{n}} \right)} \right]}^n}} $
    If that is correct then the ratio test is a better way to prove convergence.
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    Re: Convergency of a series

    Quote Originally Posted by Plato View Post
    to domy7997, you have very poor penmanship. Those are clearly m's in the image you posted.
    It may serve better if you used some devise other than a mobile phone.
    Also be reminded that this is an English language website.

    Now just to be sure that the series is: $\displaystyle \sum\limits_{n = 1}^\infty {n!{{\left[ {\log \left( {1 + \frac{2}{n}} \right)} \right]}^n}} $
    If that is correct then the ratio test is a better way to prove convergence.
    I'm sorry for my penmanship...
    It's not as you wrote. It's like the image i've posted[SUM FROM 1 TO INFINITY]
    Wolfram alpha says that it converges by the comparison test, but i don't get how is it possible...
    Thanks for your time and have a nice day.Convergency of a series-screenshot_2018-12-29-02-42-24-673_com.wolfram.android.alpha.jpeg

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    Re: Convergency of a series

    Quote Originally Posted by domy7997 View Post
    It's not as you wrote.
    That is exactly equivalent to what I wrote: $\displaystyle {\left[ {\log \left( {1 + \frac{2}{n}} \right)} \right]^n} \equiv {\log ^n}\left( {1 + \frac{2}{n}} \right)$
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  7. #7
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    Re: Convergency of a series

    we could use

    $\displaystyle \log \left(1+\frac{2}{n}\right)<\frac{2}{n}$

    $\displaystyle n!\left(\log \left(1+\frac{2}{n}\right)\right)^n<\frac{2^nn!}{n ^n}$

    and compare with the series

    $\displaystyle \sum _{n=1}^{\infty } \frac{2^nn!}{n^n}$

    which converges by the $n^{th}$ root test using

    $\displaystyle \lim_{n\to \infty } \left(\frac{\sqrt[n]{n!}}{n}\right)=\frac{1}{e}$
    Thanks from domy7997 and topsquark
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  8. #8
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    Re: Convergency of a series

    Quote Originally Posted by Idea View Post
    we could use

    $\displaystyle \log \left(1+\frac{2}{n}\right)<\frac{2}{n}$

    $\displaystyle n!\left(\log \left(1+\frac{2}{n}\right)\right)^n<\frac{2^nn!}{n ^n}$

    and compare with the series

    $\displaystyle \sum _{n=1}^{\infty } \frac{2^nn!}{n^n}$

    which converges by the $n^{th}$ root test using

    $\displaystyle \lim_{n\to \infty } \left(\frac{\sqrt[n]{n!}}{n}\right)=\frac{1}{e}$
    Yeah. Thanks... Just one think... How can the last limit can be equal to 1/e?
    Which formula for limits do you refer to?
    Thanks for your time

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    Re: Convergency of a series

    Quote Originally Posted by domy7997 View Post
    Yeah. Thanks... Just one think... How can the last limit can be equal to 1/e?
    Which formula for limits do you refer to?
    The proof is over a page long when done right. So all I will do is a brief outline.
    $\displaystyle \log \left[ {(N - 1)!} \right] \le \sum\limits_{k = 2}^N {\log (k - 1)} \le \int_1^N {\log (x)} = N\log (N) - N + 1$
    Raise to power of $e$
    $\displaystyle \begin{align*}(N-1)!&\le \dfrac{N!}{e^N}e \le N!\\\dfrac{e^N}{N}&\le\dfrac{N^N}{N!}e\le e^N\\\dfrac{e}{\sqrt[N]{N}}&\le\sqrt[N]{\dfrac{N^N}{N!}}\sqrt[N]e\le e\end{align*}$Now if $N\to\infty$ then if follow that $\sqrt[N]{\dfrac{N^N}{N!}}\to e$.

    Just take the reciprocal.
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