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this converges to zero since eventually n! expands into n*....n-(n-1) [i.e. 1], while the last multiple of n^n is still n. i.e. once n is big enough the last divisor is obviously bigger enough to bring the number to zero
hey sandra you dont have a better reason?
i dont buy this argument
how about consider x! / x^x
for your father
what you wrote isnt obvious , sorry
we can make x! continuous using gamma function
hmmm, not sure about that =) can you explain?
well you didnt show it converges
yeah I know, I gave the best reasoning I could heh
Did you guys try the ratio test?
i for one did not :p
ratio test is for series, this is a sequence
or maybe im wrong, i havent see ratio test for sequences
how about a sandwich theorem
n*n >= n * n-1 n*n*n > = n * n-1 n - 2
so by induction we have
n * n-1 * n-2 / n*n*n < = 1
it is strictly less than 1
n* n > (n-1)(n-2) for n >= 1
ok so divide out the first n
ok, spolier alert... this is a bit beyond my current understanding, but I think the by induction track sounded good... http://www.physicsforums.com/showthread.php?t=195508
the root solution is way over my head
sandra i made you a hero
omg lol! yay =)
I'm already your fan, so not much I can do :p
How do you give points to people?
you "become their fan"
I think they should add per answer type ratings
at least in addition
something like "best answer", or "thanks" or something. one more thing to collect :p
I can't find "become their fan"
this is a tough question
see , the series 1/n diverges, but the sequence 1/n converges to zero when you take the limit
hmmm jkwon, should be right by the person's name , e.g. right above my reply HERE
\[x + x = 2xx\]
we know n! > n-1 * n-2
i think we can use the fact that its less than 1 and decreasing