# Overapproximating $\binom{n/t + m}{m}^t$ [closed]

It is well-known that

$$\left(1+\frac{n}{t}\right)^t\ \le\ e^n$$

for positive integers $$n$$ and $$t$$.

You can also write the basis on the LHS in a more cumbersome way as

$$1+\frac{n}{t}\quad =\quad \binom{\frac{n}{t}+m}{m}\quad\text{for}\ m=1$$

Here, you have to somehow generalize the binomial coefficients to accept rational numbers in the upper position, e.g.:

$$\binom{x}{m}\ ≝\ \prod_{0\le i < m} \frac{x-i}{i+1}\quad\text{for}\ x\in ℚ\ \text{and}\ m\in ℕ_{\ge 0}$$ with the convention that the empty product evaluates to 1.

Now, is it possible to generalize the inequality

$$\binom{\frac{n}{t}+m}{m}^t\ \le\ e^n\quad\text{for}\ m=1$$

from $$m=1$$ to arbitrary positive integers $$m$$ such that the new upper bound (i.e., the new RHS), whatever it might be, would be elementary, would not involve the binomial coefficients (and possibly not involve the factorials), and would simplify to $$e^n$$ (i.e., the old RHS) for $$m=1$$? Of course, we'd like to have the upper bound on the generalization to be reasonably low. (I admit that the term "reasonably" is diffuse; my apologies.)

Literature references are also welcome.

## closed as off-topic by user44191, Brendan McKay, Sean Lawton, Alexey Ustinov, Boris BukhApr 16 at 12:21

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• This isn't true even for $m = 2$; the actual limit isn't difficult to determine, using the product form of the binomial coefficient. – user44191 Apr 16 at 2:42
• $(m^m/m!)^t e^n$ is elementary. Note that $m^m/m!=1$ for $m=1$. But this problem is not of MO nature. – Brendan McKay Apr 16 at 2:55
• @user49915 Sorry, I slightly misread the question itself. I do stand behind the rest of my comment, which Max Alekseyev's answer illustrates. – user44191 Apr 16 at 19:14

Noticing that $$\binom{\frac{n}{t}+m}{m}=\prod_{i=1}^m \left(\frac{n}{ti}+1\right),$$ we get $$\binom{\frac{n}{t}+m}{m}^t\leq e^{nH_m},$$ where $$H_m = \frac{1}{1}+\frac{1}{2}+\dotsb+\frac{1}{m}$$ is the $$m$$th harmonic number.