If $\lambda_n = \int_{0}^{1} \frac{dt}{(1+t)^n}$, for $n \in \mathbb{N}$, then $\,\lim_{n \to \infty} (\lambda_{n})^{1/n}=1.$

Question

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.

Answer 1

Actually, $$ \int_0^1 \frac{dt}{(1+t)^n}=\left.\frac{1}{1-n}\frac{1}{(1+t)^{n-1}}\,\right|_0^1=\frac{1}{n-1}-\frac{2^{-n+1}}{n-1} $$ and hence, for all $n>1$ $$ \frac{1}{2(n-1)}<\int_0^1 \frac{dt}{(1+t)^n}<\frac{1}{n-1}. $$ Next, observe that $$ \lim_{n\to\infty}\left(\frac{1}{2(n-1)}\right)^{1/n}=\lim_{n\to\infty}\left(\frac{1}{n-1}\right)^{1/n}=1. $$

Answer 2

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.