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Let $c > 0$. I'm trying to show that the sequence $\displaystyle\sum\limits_{k=0}^n \left|\frac{n^k-\frac{n!}{(n-k)!}}{n^k}\right|\frac{c^k}{k!}$ converges to zero, as $n \to \infty$.

I know that the nominator is a polynomial of $n$ of degree $k-1$, while the denominator is a polynomial of degree $k$. Therefore the whole expression in the absolute value behaves like $\frac{1}{n}$ for large values of $n$, and the sequence $\sum\limits_{k=0}^n \frac{1}{n}\frac{c^k}{k!}$ obviously converges to zero (it's $\frac{1}{n}$ times the expression for exponent). Therefore my problem is showing that the expression behaves like $\frac{1}{n}$. That's mainly because the polynomial in the nominator's coefficients depend on $k$.

Anybody got an idea?

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up vote 2 down vote accepted

We have $n^k-\frac{n!}{(n-k)!}=\prod_{j=1}^kn-\prod_{j=1}^k(n-j+1)$. We use the following lemma, which can be shown by induction:

If $a_k,b_k, 1\leq k\leq N$ are complex numbers of modulus $\leq R$ then $$\left|\prod_{j=1}^Na_j-\prod_{j=1}^Nb_j\right|\leq r^{N-1}\sum_{j=1}^N|a_j-b_j|.$$

We get that $$\left|n^k-\frac{n!}{(n-k)!}\right|\leq n^{k-1}\sum_{j=1}^k(j-1)=n^{k-1}\frac{k(k-1)}2$$ hence $$\sum_{k=0}^n\left|n^k-\frac{n!}{(n-k)!}\right|\frac 1{n^k}\frac{c^k}{k!}\leq \frac 1n\sum_{k=2}^n\frac{c^k}{(k-2)!}\leq \frac{c^2}ne^c.$$

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That's brilliant. I think I can also prove the lemma. Could you maybe give me a hint as for how you came up with this great idea? – George Jun 2 '12 at 14:22
Comparing the modulus of two product is used in a proof of central limit theorem. It tried to apply this idea here. Did you also post it at math.overflow ? – Davide Giraudo Jun 2 '12 at 14:23
Maybe I should've taken probability first, then... :-) I didn't, but I've been working on this idea with a friend so he might have done it. – George Jun 2 '12 at 14:27

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