integration - Evaluating $lim_{nto infty}sum_{i=1}^n frac{1}{(i+1)(i+2)}$ through Riemann sum

I am stuck in this problem: $$\lim_{n\to \infty}\sum_{i=1}^n \frac{1}{(i+1)(i+2)}$$
While it can be easily solved using partial fractions, I wanted to solve this through Riemann sums which I terribly failed at. The results were different. Here's my procedure. Please tell me where I've gone wrong.

Partial Fraction Method
$$\lim_{n\to \infty}\sum_{i=1}^n \frac{1}{(i+1)(i+2)} = \lim_{n\to \infty}\sum_{i=1}^n \frac{1}{i+1}-\frac{1}{i+2} = \lim_{n\to \infty}\frac{1}{2}-\frac{1}{n} = \frac{1}{2}$$

Reimann Sum Method

$$\lim_{n\to \infty}\sum_{i=1}^n \frac{1}{(i+1)(i+2)} = \lim_{n\to \infty}n\cdot\frac{1}{n}\cdot\sum_{i=1}^n \frac{1}{(i+1)(i+2)}$$
$\text{Let }x=\dfrac{i}{n} \implies dx=\dfrac{1}{n} \text{. Therefore the above expression reduces to:}$
$$\lim_{n\to \infty}n\int_0^1{\frac{dx}{(nx+1)(nx+2)}}$$
$\text{Supposing } t=\dfrac{nx+2}{nx+1} \text{, reduces the above expression to:}$
$$\lim_{n\to \infty}n\int_2^{\bigl(\frac{n+2}{n+1}\bigr)}{\frac{(nx+1)^2dt}{-n(nx+1)(nx+2)}} = \lim_{n\to \infty}\int_{\bigl(\frac{n+2}{n+1}\bigr)}^2{\frac{dt}{t}}=\lim_{n\to \infty}\Biggl(\ln2-\ln{\biggl(\dfrac{n+2}{n+1}\biggr)}\Biggr)=\ln2$$

Answer

When you have stated $\lim_{n\to \infty}n\cdot\frac{1}{n}\cdot\sum_{i=1}^n \frac{1}{(i+1)(i+2)}=\lim_{n\to \infty}n\int_0^1{\frac{dx}{(nx+1)(nx+2)}}$ you are using this: $$\lim_{n\to \infty}f\cdot g=\lim_{n\to \infty}f\cdot \lim_{n\to \infty}g$$ As, while converting the the sum into integral you are taking limit. But, to hold this property both of the limits need to be hold. Here, $\lim_{n\to \infty}n$ does't exists.