<s>[INST] If $\lambda_{n} = \int_{0}^{1} \frac{dt}{(1+t)^n}$ , then value of $\lim_{n \to \infty} \lambda_{n}^{1/n}$? 
Let $$\lambda_n = \int_0^1\frac{dt}{(1+t)^n} \quad \forall n \in \Bbb{N}.$$

To evaluate the integral,
Let $1+t = u$ then $dt  =du$, when $t =0$ we get $u=1$ and when $t = 1$ we get $u = 2$; so that $\lambda_{n} = \int_{1}^{2} \frac{du}{u^n}$ we get after solving $\lambda_{n} = \frac{2^{1-n} - 1}{1-n}$ for $n \neq 1$.
For $n=1$, $\lambda_{1} = \int_{0}^{1}\frac{dt}{1+t} = ln2$ and hence I think $\lambda_{n}$ exists for all $n \in \Bbb{N}$.
In order to prove that the sequence $\lambda_{n}$ is bounded, I thought if I would prove that if the sequeences are converging then it will be bounded. So checking for convergence of $\lambda_{n}$,
that is $\lim_{n \rightarrow \infty}\lambda_{n} = \lim_{n \rightarrow \infty} \frac{2^{1-n}-1}{1-n}$ , so as $n \rightarrow \infty, \lambda_{n} \rightarrow 0$ and hence as the sequence is convergent we get that the sequence of $\lambda_{n} $ are bounded. Is this correct?.
Next I have to check for the correctness of $\lim_{n \rightarrow \infty} \lambda_{n}^{\frac{1}{n}} = 1$.
$\lim_{n \rightarrow \infty} \lambda_{n}^{\frac{1}{n}} = \lim_{n \rightarrow \infty} (\frac{2^{1-n}}{1-n} - \frac{1}{1-n})^{\frac{1}{n}}$, it is coming $0^0$ indeterminate form.
So let $L =\lim_{n \rightarrow \infty}(\frac{2^{1-n}}{1-n} - \frac{1}{1-n})^{\frac{1}{n}}$.
$ln(L) = \lim_{n \rightarrow \infty}(\frac{1}{n} ln(\frac{2^{1-n}}{1-n} - \frac{1}{1-n}))$.
So next I think I would apply $\lim f.g = \lim f . \lim g$, so that I would be getting $ln(L) = \lim_{n \rightarrow \infty} \frac{1}{n} . \lim_{n \rightarrow \infty} ln(\frac{2^{1-n}}{1-n} - \frac{1}{1-n}) = \lim_{n \rightarrow \infty} \frac{1}{n} . ln (\lim_{n \rightarrow \infty} (\frac{2^{1-n}}{1-n} - \frac{1}{1-n}))$, now I am thinking how to evaluate the limit $ln (\lim_{n \rightarrow \infty} (\frac{2^{1-n}}{1-n} - \frac{1}{1-n}))$  as its of the form $ln(0)$?.
 [/INST] $$\lim_{n \to \infty} \frac{1}{n} \log \frac{1-2^{-(n-1)}}{n-1} = \lim_{n \to \infty} \frac{1}{n} \log(1-2^{-(n-1)}) - \lim_{n \to \infty} \frac{1}{n} \log(n-1) = 0,$$
so $\lambda_n^{1/n} \to 1$.
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