Evaluate $\lim_{n\to \infty}\sqrt[n]{\frac{(17n)!}{(n!)^{17}}}$

Question

Evaluate $$\lim_{n\to \infty}\sqrt[n]{\frac{(17n)!}{(n!)^{17}}}$$

I know that if we can show that $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=L$ then $\lim_{n\to\infty}\sqrt[n]{a_n}=L$

So we have to look at $$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{\sqrt[n+1]{\frac{[17(n+1)]!}{[(n+1)!]^{17}}}}{\sqrt[n]{\frac{(17n)!}{(n!)^{17}}}}$$

I managed to continue to:$$\lim_{n\to \infty}\frac{\sqrt[n+1]{\frac{(n+1)[17n]!}{(n+1)^{17}[(n!)]^{17}}}}{\sqrt[n]{\frac{(17n)!}{(n!)^{17}}}}$$

So it is of the form: $$\lim_{n\to \infty}\frac{\sqrt[n+1]{\frac{(n+1)a}{(n+1)^{17}b}}}{\sqrt[n]{\frac{a}{b}}}=\lim_{n\to \infty}\frac{\sqrt[n+1]{\frac{(n+1)}{(n+1)^{17}}}\sqrt[n+1]{\frac{a}{b}}}{\sqrt[n]{\frac{a}{b}}}=\lim_{n\to \infty}\frac{\sqrt[n+1]{\frac{(n+1)}{(n+1)^{17}}}\sqrt[n+1]{\frac{a}{b}}}{\sqrt[n]{\frac{a}{b}}}=\lim_{n\to \infty}{\sqrt[n+1]{\frac{(n+1)}{(n+1)^{17}}}\sqrt[n(n+1)]{\frac{a}{b}}}$$

But I can see how to proceed

Answer 1

You wrote: $ \frac{a_{n+1}}{a_n}=\frac{\sqrt[n+1]{\frac{[17(n+1)]!}{[(n+1)!]^{17}}}}{\sqrt[n]{\frac{(17n)!}{(n!)^{17}}}}.$

But this is not true, since $a_n= \frac{(17n)!}{(n!)^{17}}$.

With this you should get (the computations are your turn):

$ \frac{a_{n+1}}{a_n}= \frac{17n+1}{n+1} \cdot \frac{17n+2}{n+1} \cdot .... \cdot \frac{17n+17}{n+1}.$

Thus $ \frac{a_{n+1}}{a_n} \to 17^{17}$ as $n \to \infty.$