I'm having trouble finding the sublimits of the sequence:
$b_n= n!^{−1/n}$
I tried to use the even and odd terms but I don't think it's the right way to do it... How can I find the sublimits of this sequence?
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Stirling's formula? $n!\sim \sqrt {2\pi n} \left( \frac ne \right)^n$ seems helpful. – lulu Apr 06 '16 at 17:46
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Isn't there another way? – Granger Obliviate Apr 06 '16 at 17:54
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There are some other arguments here: http://math.stackexchange.com/questions/653650/infinite-series-of-nth-root-of-n-factorial – lulu Apr 06 '16 at 17:58
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Remember that
$$\lim_{n\to\infty}a_n=L\implies \lim_{n\to\infty}\frac{a_1+\ldots+a_n}n=L$$
And using the Arithmetic-Geometric-Harmonic Progressions inequality (for positive sequences):
$$L=\frac1{\frac1L}\xleftarrow[\infty\leftarrow n]{}\frac n{\frac1{a_1}+\ldots+\frac1{a_n}}\le\sqrt[n]{a_1\cdot\ldots\cdot a_n}\le\frac{a_1+\ldots+a_n}n\xrightarrow[n\to\infty]{}L$$
Apply the above to the sequence $\;\{1,2,...,n,...\}\;$ and get your sequence only has one (partial or whatever) limit: zero.
DonAntonio
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