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I have a problem calculating the following limit:

$\lim\limits_{n \to \infty}{ (1-2/3)^{3/n}*(1-2/4)^{4/n}...(1-2/(n+2))^\frac{n+2}{n}}$

I thought this is a geometric average of the first n items of a series and so I figured the limit should be the same as the limit of the infinity series: $$a_n=(1-2/(n+2))^{n+2}$$ which I though should be zero as n approaches infinity, since $(1-2/(n+2))<1$.

I would greatly appreciate if anyone could help me understand this limit.

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    $\begingroup$ As $n\to \infty$ your $a_n \to e^{-2}$. $\endgroup$
    – Joel
    Apr 23, 2015 at 16:57
  • $\begingroup$ Oh of course. Thank you. $\endgroup$
    – Ron
    Apr 23, 2015 at 16:59
  • $\begingroup$ So then, if $b_n$ are positive and $b_n \to L,$ you want to show $(b_1\cdot b_2 \cdot \dots \cdot b_n)^{1/n} \to L.$ $\endgroup$
    – zhw.
    Apr 23, 2015 at 17:15

1 Answer 1

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I will call your sequence $A_n$. If we take the logarithm of your sequence we find:

$$\lim_{n\to\infty} \ln A_n = \lim_{n\to \infty} \frac1n \left(\sum_{k=1}^{n} (k+2)\ln\left(1-\frac{2}{k+2}\right) \right).$$

If a sequence $a_n$ converges, then its Cesaro mean converges to the same value: $$\lim_{n\to \infty} a_n = \lim_{n\to \infty} \frac1n \sum_{k=1}^n a_k.$$

So we wish to show that $(k+2)\ln\left(1-\frac{2}{k+2}\right)$ converges. This is the same as the limit: $$\lim_{x\to \infty} \frac{\ln(1-2/(x+2))}{1/(x+2)}$$ and we can use l'Hopital's rule to evaluate it.

$$\lim_{x\to \infty} \frac{\ln(1-2/(x+2))}{1/(x+2)} = \lim_{x\to\infty} \frac{\frac{1}{1-2/(x+2)}\cdot \frac{2}{(x+2)^2}}{-(x+2)^{-2}} = \lim_{x\to\infty} \frac{-2}{1-2/(x+2)}=-2.$$

Therefore the logarithm of the sequence converges to $-2$, and so the original sequence converges to $e^{-2}$, as you suspected.

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  • $\begingroup$ You can find a proof of the Cesaro mean here math.stackexchange.com/questions/565288/… $\endgroup$
    – Joel
    Apr 23, 2015 at 17:19
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    $\begingroup$ I didn't study l'Hospital yet, so I proved it in a different way: $\endgroup$
    – Ron
    Apr 23, 2015 at 18:55
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    $\begingroup$ $(1-2/(n+2))^{n+2}=((1+1/((n+2/-2)))^{((n+2)/-2)})^{-2} = e^{-2}$ and then compared this result to the Geometric average of the first n items, which was the requested limit. $\endgroup$
    – Ron
    Apr 23, 2015 at 19:07
  • $\begingroup$ I am glad you were able to work it out. I just chose the quickest way I could think of to prove it using techniques from undergrad analysis, but there are other ways to organize the proof. $\endgroup$
    – Joel
    Apr 23, 2015 at 20:10

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