I tried converting the limit to a_n form, but couldn't get anywhere. Please provide help.

Question

Recall n!=n(n-1)(n-2)...3x2x1. Define the sequence {a_n} by a_n=n.(n!)^{$\frac{-1}{n}$}. Note that the lim a_n=e.

Find the value of $\displaystyle{\lim_{n \to \infty}}$ ln(2n+1).((2n+1)!)^{$\frac{-1}{2n+1}$} + $\displaystyle{\lim_{n \to \infty}}$ ln(3n).((3n)!)^{$\frac{-1}{3n}$}

I tried to convert the given limit equation $\displaystyle{\lim_{n \to \infty}}$ ln(2n+1).((2n+1)!)^{$\frac{-1}{2n+1}$} + $\displaystyle{\lim_{n \to \infty}}$ ln(3n).((3n)!)^{$\frac{-1}{3n}$} to a_n=e, but couldn't get anywhere.

Answer 1

Observe that $$\ln (n) \cdot (n!)^{-1/n} = \frac{\ln n}{n} \cdot n (n!)^{-1/n} = \frac{\ln (n)}{n} \cdot a_n$$ where both $\frac{\ln(n)}{n}$ and $a_n$ have finite limits as $n\to\infty$. Thus $$\lim_{n\to\infty} \ln (n) \cdot (n!)^{-1/n} = \left( \lim_{n\to\infty} \frac{\ln (n)}{n} \right) \left( \lim_{n\to\infty} a_n \right).$$ Now what happens if you substitute $n$ with some other expression, say $2k+1$ or $3k$, and then take $k \to \infty$?