TITLE: Calculate limit of integral sequence
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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 $1/2$.