Limit with Integration

Since $\frac{y}{1+y}<\log(1+y)<y$ We can bound the integrand as shown: $\frac{1}{x^3}<\frac{1}{nx^2\ln(1+x/n)}<\frac{1}{nx^2(x/n)/(1+x/n)}=\frac{1}{x^3}(1+x/n)$ By monotonicity, it satisfies $\color{#D61F06}{\frac{1}{2}} =\int_1^{+\infty}\frac{1}{x^3}\,\mathrm dx<\int_1^{+\infty}\frac{1}{nx^2\ln(1+x/n)}\,\mathrm dx<\int_1^{+\infty}\frac{1}{x^3}(1+x/n)\,\mathrm dx=\color{#3D99F6}{\frac{1}{2}+\frac{1}{n}}$ Take the limit and get the answer.

My first answer was incorrect because I thought $\log$ was base 10 and not base e. Perhaps use $\ln$ instead of $\log$ ?

Recall that $\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^{n} = e^x$ . Thus, the natural logarithm of this equality yields $\lim_{n \to \infty} n\ln\left(1+\frac{x}{n}\right)= x$ which, solving for $\ln\left(1+\frac{x}{n}\right)$ gives $\lim_{n \to \infty} \ln\left(1+\frac{x}{n}\right)=\frac{x}{n}$ .

We can now substitute this expression into the given integral and easily solve it.

$\displaystyle \lim_{n \to \infty} \frac{1}{n}\int_{1}^{\infty}\frac{1}{x^{2}\ln\left(1+\frac{x}{n}\right)}dx$

$\displaystyle \int_{1}^{\infty}\lim_{n \to \infty} \frac{1}{n} \frac{1}{x^{2}\ln\left(1+\frac{x}{n}\right)}dx$

$\displaystyle \int_{1}^{\infty}\frac{1}{x^{3}}dx$

$\displaystyle \lim_{a \to \infty}-\frac{1}{2a^2} + \frac{1}{2} = \boxed{\frac{1}{2}}$

Define the sequence of functions $f_n(x)$ and compute its (point-wise) limit $f(x)$ with "l'Hospital's Rule": $\begin{aligned} f_n:&& D&:=[1;\:\infty)\rightarrow\mathbb{R}, &&&&& f_n(x)&:=\frac{1}{nx^2\ln\left(1+\frac{x}{n}\right)}=\frac{1}{x^3}\cdot\frac{\frac{x}{n}}{\ln\left(1+\frac{x}{n}\right)} &&&&&\left|\lim_{n\rightarrow\infty}f_n(x)\underset{\text{l'H.}}{=}\frac{1}{x^3}=:f(x)\right. \end{aligned}$ Find a dominating function $g(x)$ using $\frac{h}{1+h}<\ln(1+h),\quad h>0\quad (*)$ : $\begin{aligned} x&>1:&&& |f_n(x)|&=f_n(x)\leq \frac{1}{x^2}\cdot\frac{1}{n\ln\left(1+\frac{1}{n}\right)}\underset{(*)}{<}\frac{1}{x^2}\left(1+\frac{1}{n}\right)\leq\frac{2}{x^2}=:g(x)\text{ integrable on }D \end{aligned}$ Interchange limit and integration using Lebesgue's Dominated Convergence Theorem : $\lim_{n\rightarrow\infty}\int_1^\infty f_n(x)\:dx=\int_1^\infty f(x)\:dx=\left[\frac{-1}{2x^2}\right]_1^{\infty}=\boxed{\frac{1}{2}}$