# Find $\lim_{n\to \infty}\int_1^{\infty}\frac{n}{1+x^n}dx$ [duplicate]

Find $$\lim_{n\to \infty}\int_1^{\infty}\frac{n}{1+x^n}dx$$

Lebesgue dominated convergence does not work. My next thought was to use Fatou's lemma and reverse Fatou's lemma to bound the integral, but I haven't the faintest idea how to calculate the liminf or limsup of the integrand. Wolfram Alpha tells me that the value of the integral decreases (starting with n=2) from about 1.2 to about 0.7, but I can't even show that the sequence is decreasing right now.

Any hints would be appreciated.

## marked as duplicate by rtybase, saz measure-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 26 at 19:17

Set $u=x^n$ to get: $$I_n:=\int^\infty_1\frac{n}{1+x^n}\,dx=\int^\infty_1 \frac{1}{u^\frac{n-1}{n}(1+u)}\,du$$ For $n\geq 2$, the inequality $u^\frac{n-1}{n}\geq u^\frac{1}{2}$ holds pointwise for $u\in [1,\infty)$ so: \begin{align} \left| \frac{1}{u^\frac{n-1}{n}(1+u)}\right|\leq \frac{1}{\sqrt[]{u}(1+u)} \end{align} Hence by dominated convergence theorem we get: \begin{align} \lim_{n\to\infty}I_n = \int^\infty_1 \lim_{n\to\infty}\frac{1}{u^\frac{n-1}{n}(1+u)}\,du=\int^\infty_1 \frac{1}{u(1+u)}\,du = \log(2) \end{align} And indeed $\log(2)\approx 0.7$.