Evaluate: $$\lim_{n\rightarrow \infty} \left[ \dfrac{1}{(n+1)(n+2)} + \dfrac{2}{(n+2)(n+4)} + \cdots + \dfrac{n}{6n^2} \right]$$
$\text{My Attempt:}$ breaking down the summation series into: $$\sum_{r=1}^{n} \dfrac{r}{(n+r)(n+2r)}$$.
Further breaking down into two separate series: $$\sum_{r=1}^{n} \dfrac{r}{(n+r)(n+2r)}=\sum_{r=1}^{n} \dfrac{(n+2r)-(n+r)}{(n+r)(n+2r)}$$
This will reduce to give: $$\sum_{r=1}^{n} \dfrac{1}{n+r} – \sum_{r=1}^{n}\dfrac{1}{n+2r}$$
Now, applying limits to the sum: $$\lim_{n\rightarrow\infty}\left[\sum_{r=1}^{n} \dfrac{1}{n+r} – \sum_{r=1}^{n} \dfrac{1}{n+2r}\right]$$
Taking $n$ common in denominator and converting to Definite integral taking $\dfrac{r}{n}=x$ this reduces to: $$\int_{0}^{1}\dfrac{\text{dx}}{1+x}-\int_{0}^{1} \dfrac{\text{dx}}{1+2x}$$
Edit: Solving this we will get the answer as $\ln\left(\dfrac{2}{\sqrt{3}}\right)$.
I have had committed an error in the evaluation of the 2nd integral as Mr. Robert Z has pointed out below.
Best Answer
After revising my answer, I got: as $n\to \infty$, $$\begin{align}\sum_{r=1}^{n} \dfrac{r}{(n+r)(n+2r)}&=\frac{1}{n}\sum_{r=1}^{n} \dfrac{\frac{r}{n}}{(1+\frac{r}{n})(1+2\frac{r}{n})}\\ &\to \int_0^1 \frac{x}{(1+x)(1+2x)}\,dx\\ &=\int_0^1\left(\frac{1}{1+x}-\frac{1}{1+2x}\right)\,dx\\ &=\left[\ln\left(\frac{1+x}{(1+2x)^{1/2}}\right)\right]_0^1\\ &= \ln(2/\sqrt{3}). \end{align}$$ So, your approach is correct, and there is just a minor error in the integration of $1/(1+2x)$.