Let $k$ and $n$ be positive integers.

Can we show the following inequality: \begin{align} \frac{ (n+k)!}{n! \sqrt{n+k}} \le ( f(n) )^k \sqrt{k!}, \end{align} where $f(n)$ is some funciton of $n$ only. For this assume that $n$ is fixed, and the above must hold for all positive integers $k$.

  • $\begingroup$ @Clayton I added a correction. $\endgroup$ – Lisa Mar 19 at 19:46
  • $\begingroup$ Do you have Stirling's Approximation? Do you have any specific thoughts about the inequality? (Hint: Focus on the fact that $n$ is fixed.) $\endgroup$ – Clayton Mar 19 at 19:53
  • $\begingroup$ @Clayton Yes, I tried Stirling's bounds. However, I can not decouple $n$ and $k$ in the sum $(n+k)$. $\endgroup$ – Lisa Mar 19 at 20:01

We cannot show the inequality asked about because the left-hand side essentially grows at a rate like $k!$ while the right-hand side grows at a rate of $c^k\sqrt{k!}$, for some $c>0$. To be more precise, we have for sufficiently large $k$, that

$$ c_n\cdot\frac{k!}{\sqrt{k}}\leq c_n\frac{(n+k)!}{\sqrt{k}}\leq\frac{(n+k)!}{n!\sqrt{n+ k}}. $$ Thus, if we show the left-hand side cannot be bounded above by $c^k\sqrt{k!}$, then we'll be done.

At this point, if the proposed inequality were true, we'd have $$ c_n\cdot\frac{\sqrt{k!}}{\sqrt{k}}\leq c^k, $$ which follows just from dividing both sides by $\sqrt{k!}$. Here, I'm using $c=f(n)$; since $n$ is fixed, it means $f(n)$ is simply a constant.

Taking $k^{th}$ roots and applying Stirling's Approximation, we deduce for $k$ sufficiently large, that $$ \frac{1}{2}\cdot\left(\frac{k}{e}\right)^{1/2}\leq c. $$ Letting $k\to\infty$, we see the left-hand side tends to $\infty$, while the right-hand side remains fixed at $c$. This is clearly a contradiction, so there is no such $f$.

Note: This problem relies on $n$ being fixed. If you allow $n$ to vary, you can prove that such a function does exist, but $n$ will necessarily depend on $k$.


Let suppose that such a $f$ exists, and fix $n$. You have then $$\frac{(n+k)!}{n! \sqrt{(n+k)k!}(f(n))^k} \leq 1$$

Now, when $k$ tends to $\infty$, you have by Stirling formula $$\frac{(n+k)!}{n! \sqrt{(n+k)k!}(f(n))^k} \sim\frac{\sqrt{2\pi(n+k)} \left(\frac{n+k}{e} \right)^{n+k}}{n! \sqrt{(n+k)}\sqrt{\sqrt{2\pi k} \left(\frac{k}{e} \right)^k}(f(n))^k} = L \frac{(n+k)^{n+k}}{e^{k/2} k^{(2k+1)/4}(f(n))^k} $$

where $L$ is a constant independant of $k$. Now, note that when $k$ tends to $\infty$, $$(n+k)^{n+k} = k^{n+k} \left( 1 + \frac{n}{k}\right)^{n+k} \sim k^{n+k} e^n $$

You deduce that $$\frac{(n+k)!}{n! \sqrt{(n+k)k!}(f(n))^k} \sim L' \frac{k^{n+k}}{e^{k/2} k^{(2k+1)/4}(f(n))^k} \sim L' \frac{k^{n- \frac{1}{4}}}{ (\sqrt{e}f(n))^k} k^{k/2}$$

and it is easy to see that the limit when $k$ tends to $+\infty$ is $+\infty$ ($k^k$ is stronger than all the other terms).

This is impossible because it should always be $\leq 1$.

So there does not exist such a function $f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.