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Let be a polynomial with real coefficients. Calculate the value of this limit:

$$\lim_{n \rightarrow \infty} |P(1)...P(n+1)|^ \frac1{n+1}-|P(1)...P(n)|^ \frac1{n} $$

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Hint: first try it with $P(x)=ax^m$, the leading term. You can later argue that the other terms don't matter as $n \to \infty$ – Ross Millikan May 25 '12 at 11:28
It seems the obvious first thing to try is to replace $P$ with its behavior "at $+\infty$": i.e. $P(x) \sim a x^k$ for some integer $k$ and real number $a$. Then see if you can compute this limit, and prove it equal to the original. Disclaimer: I haven't carried out the calculation yet. – Hurkyl May 25 '12 at 11:30
If $P$ have zero at integer points then the limit is zero. On the other hand, for $P(n)=n$ the limit is $1/e$. – Norbert May 25 '12 at 11:37
@Ross Millikan: having received these information it means that the limit seems to be $\frac{|a|}{e^{m}}$ for m=1, and infinity for m>1. I wonder how i can argue that the other terms don't matter. – user 1618033 May 25 '12 at 11:41
Note that for $P(n)=(n-1)(n-2)(n-3)$ the limit is zero – Norbert May 25 '12 at 11:43

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