Limit of definite integral sequence This is taken from an admission exam textbook at our local construction university :
$$ \lim_{n\to\infty} n\int_{1}^{2} \frac{1}{x^2(1+x^n)}dx = ?$$
I tried finding tight bounds in order to use the sandwich theorem, but that got me nowhere. I also tried using Lebesgue's Dominated Convergence Theorem, but I ended up with nothing once more.
Do you have any ideas?
 A: Using 
\begin{align}
\frac{1}{1+x^{n}} = \frac{1}{x^{n}} \, \frac{1}{1 + \frac{1}{x^{n}}} = \frac{1}{x^{n}} \, \sum_{k=0}^{\infty} \frac{(-1)^{k}}{x^{kn}}
\end{align}
then
\begin{align}
I_{n} &= \int_{1}^{2} \frac{dx}{x^{2} \, (1+x^{n})} \\
&= \int_{1}^{2} \frac{1}{x^{n+2}} \, \left( 1 - \frac{1}{x^{2n+2}} + \cdots \right) \, dx \\
&= - \frac{1}{n+1} \left[ \frac{1}{x^{n+1}} \right]_{1}^{2} + \frac{1}{2n+1} \, \left[ \frac{1}{x^{2n+1}} \right]_{1}^{2} + \cdots \\
&= \frac{1}{n+1} \, \left(1  - \frac{1}{2^{n+1}} \right) + \frac{1}{2n+1} \, \left( \frac{1}{2^{2n+1}} - 1 \right) + \cdots \\
&= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{nk+n+1} \, \left( 1 - \frac{1}{2^{nk+n+1}} \right)
\end{align}
Now,
\begin{align}
\lim_{n\to\infty} \left\{ n \, I_{n} \right\} = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+1} = \ln 2. 
\end{align}
A: Let $x=y^{1/n}.$ Then the expression equals
$$\int_1^{2^n}\frac{dy}{y^{1+1/n}(1+y)}.$$
A straightforward dominated convergence argument then shows the limit of the above is
$$\int_1^{\infty}\frac{dy}{y(1+y)} = \ln 2.$$
A: Let $f_n(x)$ be the sequence of functions on $[1,2]$ given by 
$$f_n(x)=\frac{n}{x^2(1+x^n)}$$
Furthermore, let $I(n)$ denote the integral of $f_n(x)$ 
$$\begin{align}
I(n)&=\int_1^2f_n(x)\,dx\\\\
&=\int_1^2\frac{n}{x^2(1+x^n)}dx
\end{align}$$
We will evaluate the limit $\lim_{n\to \infty}I(n)$.

First, we observe that for $1<x\le2$, 
$$\lim_{n\to\infty}\left(\frac{n}{x^2(1+x^n)}\right)=0$$
but that $f_n(1) \to \infty$.  Thus, all of the "action" of the integral occurs in a small neighborhood of $x=1$.  With that in mind, we fix a number $\delta >0$, and split $I(n)$ as follows:
$$\begin{align}
I(n)&=\int_1^2\frac{n}{x^2(1+x^n)}dx\\\\
&=\int_1^{1+\delta}\frac{n}{x^2(1+x^n)}dx+\int_{1+\delta}^2\frac{n}{x^2(1+x^n)}dx \tag 1
\end{align}$$
Note that the second integral in $(1)$ goes to zero as $n$ goes to infinity.  We now proceed to evaluate the first integral.
$$\begin{align}
\int_{1}^{1+\delta}\frac{n}{x^2(1+x^n)}dx&=\int_2^{1+(1+\delta)^n}\frac{(x-1)^{-1-1/n}}{x}dx\tag 2\\\\
&=\int_{\log 2}^{\log(1+(1+\delta)^n)}(e^x-1)^{-1-1/n}dx \tag 3\\\\
&\to \int_{\log 2}^{\infty}\frac{1}{e^x-1}dx\tag 4\\\\
&=\left.\log(1-e^{-x})\right|_{\log 2}^{\infty}\\\\
&=\log 2
\end{align}$$
Putting it all together, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\int_1^2\frac{n}{x^2(1+x^n)}dx=\log 2}$$ 
which agrees with the result reported by @Leucippus!
In arriving at $(2)$, we substituted $x\to (x-1)^{1/n}$.
In going from $(2)$ to $(3)$, we substituted $x\to e^x$.
In going from $(3)$ to $(4)$, we noted that as $n\to \infty$, the upper integration limit approaches $\infty$ and $(e^x-1)^{-1+1/n} \to (e^x-1)^{-1}$
