I'm trying to find and prove the value oflimn→∞1n(n!)1n
I was thinking that since
1n(n!)1n=1n[(1)1n(2)1n...(n)1n]
and we know that
limn→∞n1n=1
and
k1n≤ n1n ∀k≤n
So (n!)1n≤1 ∀n * Then it is bounded. We also know that lim1n=0, therefore
limn→∞1n(n!)1n=0∗∗
I'm pretty sure this line of reasoning is ok. Now proving it is another thing. Any suggestions?
*I realize now that this statement is not true, but that did not get me any closer to solving the problem. I do believe that it is bounded though.
**LOL, I don't believe this to be true anymore either, a wild guess tells me that the solution may be 1e
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