Limits And Summations!

Since

$\displaystyle \lim_{n \to \infty} \ln \left( \sqrt[n]{\dfrac{1}{n} \cdot \dfrac{2}{n} \cdots \dfrac{n}{n} } \right) = \lim_{n \to \infty} \dfrac{1}{n} \left( \sum_{k=1}^n \ln \left( \dfrac{k}{n} \right) \right) = \int_0^1 \ln (x) \ \mathrm{d}x = -1$

We have $\dfrac{1}{\sqrt[n]{n!}} \sim \dfrac{e}{n}$ as $n \to \infty$ . Thus, it suffices to prove that $\displaystyle \log_a \left( \sum_{k=1}^{a^n} (1+k)^{a^{-n}} \right) \sim n$ as $n \to \infty$ .

Now,

$a^n \cdot 2^{1/a^n} \leq \sum_{k=1}^{a^n} (1+k)^{a^{-n}} \leq a^n (1+a^n)^{1/a^n}$

and therefore using the fact that $\log_a$ is increasing,

$n + \dfrac{\log_a (2)}{a^n} \leq \log_a \left( \sum_{k=1}^{a^n} (1+k)^{a^{-n}} \right) \leq n + \dfrac{n}{a^n} + \dfrac{1}{a^n} \log_a \left( 1 + \dfrac{1}{a^n} \right)$ .

The squeeze principle now shows that

$\large \lim_{n \to \infty} \dfrac{1}{n} \log_a \left( \displaystyle \sum_{k=1}^{a^n} (1+k)^{a^{-n}} \right) = 1$

And thus we obtain $L = e \approx 2.718$