Prove that $ \left\{\frac{\binom{n+k}{k} }{(n+k)^k}\right\}_{n=1}^{\infty} $ converges to $\frac{1}{k!}$.

Question

Prove that $$ \left\{\frac{\binom{n+k}{k}}{(n+k)^k}\right\}_{n=1}^{\infty} $$ converges to $\frac{1}{k!}$, where $$\binom{n+k}{k} =\frac{(n+k)!}{n!k!}. $$

I'm not even sure how to approach this problem. Here is the scratch work that I have so far:

$$ \begin{aligned} \left|\frac{\begin{pmatrix} n+k\\ k \end{pmatrix}}{(n+k)^k}-\frac{1}{k!}\right| &=\left|\frac{(n+k)!}{n!k!(n+k)^k}-\frac{1}{k!}\right|\\ &=\left|\frac{(n+k)!-n!(n+k)^k}{n!k!(n+k)^k}\right|\\ &=\left|\frac{(n+k)[((n+k)-1)!-n!(n+k)^{k-1}]}{n!k!(n+k)^k}\right|\\ &=\left|\frac{((n+k)-1)!-n!(n+k)^{k-1}}{n!k!(n+k)^{k-1}}\right| \end{aligned} $$

I'm guessing that when I choose an epsilon, it'll probably have $k!$ in it somewhere. Please help me with this problem. Thank you in advance for your assistance.

Answer 1

Remember that $\dbinom{n+k}{k} = \dfrac{(n+k)!}{k!n!} = \dfrac{(n+k)(n+k-1)\cdots(n+1)}{k!}$.

Therefore, $\dfrac{\dbinom{n+k}{k}}{(n+k)^k} = \dfrac{1}{k!} \cdot \dfrac{n+k}{n+k} \cdot \dfrac{n+k-1}{n+k} \cdots \dfrac{n+1}{n+k}$.

Do you see where to go from here?