Calculate limit of integral sequence Hi i need to calculate limit of integral sequence:
$$\lim_{n\rightarrow\infty}\frac{1}{n}\int_{1}^{\infty}\frac{dx}{x^2\cdot\ln{(1+\frac{x}{n})}}=\lim_{n\rightarrow\infty}\int_{1}^{\infty}\frac{dx}{x^2\cdot n\cdot\ln{(1+\frac{x}{n})}}=\int_{1}^{\infty}\frac{dx}{x^2\cdot e^x}$$ and it's true if i can use monotone convergence theorem. Everything seem fine but i am not sure about monotonicity of $f_n$. I thinks it's decreasing and according to statement it should be otherwise. And taking negative doesn't seem right at all.
And another one.
$$\lim_{n\rightarrow\infty}\int_{0}^{n}(1+\frac{x}{n})^{n+1}\cdot e^{-2x}=\int_{0}^{\infty}e^{-x}$$ and here i am not sure about the $\infty$ in integral. 
I am hopeless newbie so i will be glad for thorough explanation.
 A: If you are not familiar with dominated convergence and such things, you could use the inequality
$$
\frac{y}{1+y}<\ln(1+y)<y
$$
which yields
$$
\int_1^{+\infty}\frac{1}{x^3}\,dx<\frac{1}{n}\int_1^{+\infty}\frac{1}{x^2\ln(1+x/n)}\,dx<\int_1^{+\infty}\frac{1}{x^3}+\frac{1}{n}\frac{1}{x^2}\,dx.
$$
Calculate the integrals and use the squeeze theorem, and you will see that the limit is $1/2$.
A: Change variables in your first integral $x/n=t$, yielding
$$
\lim_{n\to\infty}\frac{1}{n}\frac{n}{n^2}\int_{1/n}^\infty \frac{dt}{t^2\ln(1+t)}=\lim_{n\to\infty}\frac{1}{n^2}\int_{1/n}^\infty \frac{dt}{t^2\ln(1+t)}\ .
$$
Then use L'Hopital and compute
$$
\lim_{n\to\infty}\frac{\partial_n \int_{1/n}^\infty \frac{dt}{t^2\ln(1+t)}}{2n}\ ,
$$
whose numerator can be evaluated using the fundamental theorem of calculus
$$
\partial_n \int_{1/n}^\infty \frac{dt}{t^2\ln(1+t)}=\frac{1}{n^2}\frac{1}{(1/n)^2\ln (1+1/n)}\ .
$$
Eventually, your limit is equivalent to computing
$$
\lim_{n\to\infty}\frac{1}{2n\ln(1+1/n)}=1/2\ .
$$
A: Note that, for $x\ge1$,
$$
\lim_{n\to\infty}\frac{1}{x^2}\frac{1}{n \ln(1+x/n)}=\frac{1}{x^2}\frac{1}{\ln(e^x)}=\frac{1}{x^3}.
$$
Furthermore such an integrand is controlled as follows
$$
\frac{1}{x^2}\frac{1}{n \ln(1+x/n)}\le \frac{1}{x^2}\frac{1}{n \ln(1+1/n)}
$$
and since $\lim[n\ln(1+1/n)]=1$, there will be a constant $C$ such that
$$
\frac{1}{x^2}\frac{1}{n \ln(1+x/n)}\le\frac{C}{x^2};
$$
in fact, it can be shown that $(1+1/n)^n$ is monotone and increasing, so that $C=1/\ln2$ is fine, by the monotonicity of the logarithm. Since the integrand has a ''guardian'' $C/x^2$ (an estimate from above) which is summable,
$$
\int_{1}^{+\infty}\frac{C}{x^2}=C,
$$
we can apply Lebesgue's dominated convergence theorem and take the limit $n\to\infty$ inside the integral sign:
$$
\lim_{n\to\infty}\int_{1}^{+\infty}\frac{1}{x^2n\ln(1+x/n)}dx=
\int_{1}^{+\infty}\lim_{n\to\infty}\frac{1}{x^2n\ln(1+x/n)}dx=\int_0^{+\infty}\frac{1}{x^3}dx=1/2.
$$ 
Similarly, you can work out the second example.
