I was given as HW to calculate: $\lim_{n \rightarrow \infty} \sqrt [n] {nk \choose n}$.
I tried to use a theorem that says: if $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=L$ then $\lim_{n \rightarrow \infty} \sqrt [n] {a_n}=L$.
It's still too complicated.
Thank you.
Using Stirling's approximation we get $$ \begin{align} {nk \choose n} &=\frac{(nk)!}{n!(nk-n)!}\\ &\approx\frac{\left(\frac{nk}{e}\right)^{nk}\sqrt{2\pi nk}}{\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}\left(\frac{nk-n}{e}\right)^{nk-n}\sqrt{2\pi}}\\ &=\frac{(nk)^{nk}}{n^{n}(nk-n)^{nk-n}}\frac{\sqrt{2\pi nk}}{\sqrt{2\pi n}\sqrt{2\pi (nk-n)}}\\ &=\frac{n^{nk}k^{nk}}{n^n n^{nk-n}(k-1)^{nk-n}}\frac{\sqrt{k}}{\sqrt{2\pi n(k-1)}}\\ &=\left(\frac{k^k}{(k-1)^{k-1}}\right)^n\sqrt{\frac{k}{k-1}}\frac{1}{\sqrt{2\pi n}}\\ \end{align} $$ From here it is easy to see that $$ \lim\limits_{n\to\infty}{nk\choose n}^{1/n}=\frac{k^k}{(k-1)^{k-1}} $$
Here are two Stirling-free proofs.
Proof 1:
Write $\binom{kn}{n}$ as a quotient of two central multinomial coefficients: $\binom{kn}{\underbrace{n,n,\cdots,n}_{k\text{ times}}} \cdot \left(\binom{(k-1)n}{\underbrace{n,n,\cdots,n}_{k-1\text{ times}}} \right)^{-1}$.
Now, to finish, it is enough to prove $\lim \binom{kn}{\underbrace{n,n,\cdots,n}_{k\text{ times}}}^{1/n} = k^k$. The proof of this is really nice: $\binom{kn}{\underbrace{n,n,\cdots,n}_{k\text{ times}}}$ is smaller than the sum $\sum_{a_1 + \cdots + a_k =nk} \binom{kn}{a_1,\cdots,a_k}=(\underbrace{1+\cdots+1}_{k\text{ times}})^{nk} = k^{nk}$, and so $\binom{kn}{\underbrace{n,n,\cdots,n}_{k\text{ times}}}^{1/n} \le k^k$.
To get a lower bound, just prove that the maximum of $f(a_1, \cdots, a_k)=\binom{nk}{a_1,a_2,\cdots,a_k} (\sum a_i = nk)$ is achieved at $\forall i: a_i = n$ by some convexity argument, and so $\binom{kn}{\underbrace{n,n,\cdots,n}_{k\text{ times}}}$ is bigger than the average of all $\binom{nk}{a_1,a_2,\cdots,a_k}$, which is $\frac{k^{kn}}{\binom{nk+k-1}{k-1}}$ (the denominator is number of solutions to $a_1 + \cdots + a_k = nk$, which is a polynomial in $n$ of degree $k-1$), and so $\binom{kn}{\underbrace{n,n,\cdots,n}_{k\text{ times}}}^{1/n} \ge \frac{k^k}{1+o(1)}$ and we're done.
Proof 2:
Lemma: $\lim \frac{n!^{1/n}}{n} = e^{-1}$. Given the lemma, $\lim \frac{(kn)!^{1/n}}{(kn)^k} = e^{-k}, \lim \frac{((k-1)n)!^{1/n}}{((k-1)n)^{k-1}} = e^{-(k-1)}$, and we find via some algebra:
$$\lim \binom{kn}{n}^{1/n} = \lim \frac{(kn)!^{1/n}}{(kn)^k} \left(\frac{((k-1)n)!^{1/n}}{((k-1)n)^{k-1}} \right)^{-1} \left( \frac{(n)!^{1/n}}{n} \right)^{-1} \frac{k^k}{(k-1)^{k-1}}= \frac{k^k}{(k-1)^{k-1}}$$
Proof of lemma: For example, prove the inequality $\frac{e^{n-1}}{n} \le \frac{n^n}{n!} \le e^{n-1}$ by induction, and take $n$'th root. This inequality can also be proved by taking logarithms and using the trivial bound $\int_{i-1}^{i} \ln x dx < \ln i < \int_{i}^{i+1} \ln x dx$.
