An intriguing 'Limits' Problem!

We have $a_n =\sqrt[n]{\displaystyle \int_0^1 \left (1 + x^n \right )^n \ dx} \leq \sqrt[n]{\displaystyle \int_0^1 2^n \ dx}$

$\Rightarrow a_n \leq 2$

Also, $\sqrt[n]{\displaystyle \int_0^1 \left (1 + x^n \right )^n \ dx} = \sqrt[n]{\displaystyle \int_0^1 \sum_{k=0}^n {n \choose k} \cdot x^{nk} \ dx}$

$=\sqrt[n]{\displaystyle \sum_{k=0}^n {n \choose k} \cdot \frac{1}{nk+1}} \geq \sqrt[n]{\displaystyle \sum_{k=0}^n {n \choose k} \cdot \frac{1}{n(k+1)}}$

Since, $\sqrt[n]{\displaystyle \sum_{k=0}^n {n \choose k} \cdot \frac{1}{n(k+1)}} = \sqrt[n]{\displaystyle \sum_{k=0}^n {n+1 \choose k+1} \cdot \frac{1}{n(n+1)}} = \sqrt[n]{\dfrac{2^{n+1}-1}{n(n+1)}}$

$\Rightarrow \sqrt[n]{\displaystyle \int_0^1 \sum_{k=0}^n {n \choose k} \cdot x^{nk} \ dx} \geq \sqrt[n]{\dfrac{2^{n+1}-1}{n(n+1)}}$

So, we get $\sqrt[n]{\dfrac{2^{n+1}-1}{n(n+1)}} \leq a_n \leq 2$

Or $\displaystyle \lim_{n \to \infty} \sqrt[n]{\frac{2^{n+1}-1}{n(n+1)}} \leq \displaystyle \lim_{n \to \infty} a_n \leq 2$ .

But, $\displaystyle \lim_{n \to \infty} \sqrt[n]{\dfrac{2^{n+1}-1}{n(n+1)}} = 2$

$\Rightarrow 2 \leq \displaystyle \lim_{n \to \infty} a_n \leq 2$

Therefore $\large{\lim_{n \to \infty} \sqrt[n]{\int_0^1 \left (1 + x^n \right )^n \ dx}= \boxed{2}}$

Expanding the integrand we get $\sum_{k=0}^{n} \binom{n}{k}(x^n)^{n-k}$

So $\int_0^1 (1+x^n)^n dx = \sum_{k=0}^n \int_0^1 \binom{n}{k} (x^n)^{n-k} dx = \sum_{k=0}^n \binom{n}{k} \int_0^1 (x^n)^{n-k} dx = \sum_{k=0}^n \binom{n}{k} \cdot 1= 2^n$

Hence $\lim_{n \to \infty} \sqrt[n] {\int_0^1 (1+x^n)^n dx} = \lim_{n\to\infty} \sqrt[n] {2^n} = 2$