2

Consider the following:

$$ \Bigg| \dfrac{\prod_{i=0}^{k-1} (n-i) }{n^k} - 1 \Bigg| \leq \frac{C}{n}, \forall k \leq n $$

How to find an expression for $C$ independent of $k$ and thus $n$? It arises in estimating the speed of convergence of exponential series and was mentioned in here for example. The suggestion there was to use the triangle inequality. It is clear that $n^k$ would cancel in the numerator, but the rest still seems to be something which grows with $k$. Any ideas or thoughts?

A numeric investigation of some terms shows that there is something like an exponential or geometric series which should converge. There needs to be an upper bound on such expressions independent from $n$.

enter image description here

Community
  • 1
Rubi Shnol
  • 1,125
  • 8
  • 28
  • For fixed $k$, we get $$1 - \prod_{i = 1}^{k-1} \biggl(1 - \frac{i}{n}\biggr) = \frac{k(k-1)}{2n} + O(n^{-2}),$$ so you need $C \geqslant \frac{k(k-1)}{2}$ to have that estimate for all $n$ (for small $n$, that may not suffice, so it's just a lower bound). Thus you can't have one $C$ that works for all $n,k$ such that $k \leqslant n$. – Daniel Fischer Jul 30 '15 at 13:29
  • And it seems to be a bit counter-intuitive for me. As described in the referenced question, this term needs to go to zero as $n$ goes to infinity. But $k$ depends on $n$. It may also be equal to $n$. – Rubi Shnol Jul 30 '15 at 13:34
  • Every $k$ has its own $C$ there, perhaps it would have been better to write $\frac{C_k}{n}$ to make the dependence clear. However, note also that the whole thing is bounded by $1$. You use the bound $1$ to cut off at a certain $r$ - the remainder is then bounded by $$\sum_{m = r+1}^\infty \frac{A^m}{m!},$$ which can be made arbitrarily small by choosing $r$ large enough. Then for $k \leqslant r$ you use the bound $\frac{C_k}{n}$ to show that for large enough $n$ the sum of the first terms is small enough. – Daniel Fischer Jul 30 '15 at 14:04

1 Answers1

1

For $k=n$, the inequality becomes $$ n\left|\,\frac{n!}{n^n}-1\,\right|\le C $$ Unfortunately, $$ \lim_{n\to\infty}n\left|\,\frac{n!}{n^n}-1\,\right|=\infty $$ Thus, there cannot be a single $C$ for all $k$ and $n$.

However, this does not invalidate proofs as given in the cited answer because for $n$ big enough, $\frac{x^n}{n!}$ is small. Breaking the sum into parts allows things to work. This is done in this answer, where the sum is split into $0\le k\lt N$ and $k\ge N$.

Community
  • 1
robjohn
  • 345,667