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.

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
$$
\lim_{n\to\infty}\left(\frac{1}{2(n-1)}\right)^{1/n}=\lim_{n\to\infty}\left(\frac{1}{n-1}\right)^{1/n}=1.
$$
