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    Hey, trying to prove that there exists an epsilon in (0,e) such that there exists an N in N so the following holds;

     \displaystyle\left(\frac{n}{e-\epsilon}\right)^n < n! for all n > N.

    I think it is true (noting that (n/3)^n < n! < (n/2)^n) and then abusing stirlings formula a bit to obtain no contradictions if that was a case.

    However, I am finding it hard to prove that it does exists.
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    I'm not sure but I would rearrange to get

     \frac{n^na^n}{n!}&lt;1 letting \frac{1}{e- \epsilon}=a

    Now, think of it as a sequence. Does it converge?
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    That dosen't really do alot as far as I can tell because convergence still depends upon the value of a - for instance if a = 1/2 we have divergence and if a <= 1/3 we have convergence. I just choose 1/(e-epsilon) because of these facts but deducing anything about the behaviour with such values of a is proving hard.
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    (Original post by DeanK22)
    That dosen't really do alot as far as I can tell because convergence still depends upon the value of a - for instance if a = 1/2 we have divergence and if a <= 1/3 we have convergence. I just choose 1/(e-epsilon) because of these facts but deducing anything about the behaviour with such values of a is proving hard.
    Use the ratio lemma? To find exactly when it changes from convergence to diverge BECAUSE of a change in a.

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    \frac{a_{n+1}}{a_n} \rightarrow ea
    If ea>1 then it's gonna diverge. This happens precisely when a>e^-1, and since a=\frac{1}{e-\epsilon} a is going to be bigger than e^-1
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    Merci, never come across that lemma before but it is pretty useful!
 
 
 

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