If $\lambda_n = \int_{0}^{1} \frac{dt}{(1+t)^n}$, for $n \in \mathbb{N}$, then $\,\lim_{n \to \infty} (\lambda_{n})^{1/n}=1.$ If $\displaystyle\lambda_n = \int_{0}^{1} \frac{dt}{(1+t)^n}$ for $n \in \mathbb{N}$. Then prove that $\lim_{n \to \infty} (\lambda_{n})^{1/n}=1.$
$$\lambda_n=\int_{0}^{1} \frac{dt}{(1+t)^n}= \frac{2^{1-n}}{1-n}-\frac{1}{1-n}$$
Now if we use L'Hôpital's rule, then it gets cumbersome. Is there any short method? Thank you.

In general, if $f(x)$ is a continuous and non-negative function on $[0,1]$, 
$$ \lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}f(x)^n\,dx} = \max_{x\in[0,1]}f(x) $$
by the inequality between means.
