# Limit of a Sum $\lim_{n \to \infty }\sum_{r=1}^{n}{\frac{1}{(n+r)(n+2r)}}$

I need help in evaluating this problem: $$\lim_{n \to \infty }\sum_{r=1}^{n}{\frac{1}{(n+r)(n+2r)}}$$.

I tried to convert in to the form $$\lim_{n \to \infty }\frac{1}{n}\sum_{r=1}^{n}{f\left(\frac{r}{n}\right)}$$ to convert it into a definite integral, but it doesn't seem to be convertible to that form.

Is the answer $$0$$ because every term evaluates to zero if we apply the limit $$n \to \infty$$?

The answer given in the book where I found it is $$ln(1.5)$$, but that answer will come if there's an $$n$$ in the numerator of every term. So, I'm not sure whether it is a misprint or there is some other method which I'm not aware of.

The answer is indeed zero. Let $$f(x)=\frac{1}{(1+x)(1+2x)}$$ so that $$\int_0^1 f(x)\, \mathrm{d}x=\ln(3/2)$$. Now to prove that your limit is zero, use the rule $$\lim_{n\to\infty}\left[f(x)\cdot g(x)\right]= \left(\lim_{n\to\infty} f(x)\right)\cdot\left(\lim_{n\to\infty} g(x)\right)$$ whenever both of the limits on the right-hand side exist:
$$\lim=\lim_{n\to\infty}\left[\left(\frac{1}{n}\right)\cdot\left(\frac{1}{n}\sum_{r=1}^nf\left(\frac{r}{n}\right)\right)\right]=0\cdot\ln(3/2)=0.$$
If your book indicates that the answer is $$\ln(3/2)$$, then probably there is a misprint in either the problem or the answer. Unfortunately, misprints in textbooks (even math textbooks) happen all the time. Nicely spotted!
\begin{align} \sum_{r=1}^{n}{\frac{1}{(n+r)(n+2r)}} &<\sum_{r=1}^{n}\frac 1{n^2}\\ &=\frac 1n\\ &\xrightarrow{n\to\infty}0 \end{align}
On the other hand \begin{align} \sum_{r=1}^{n}{\frac{n}{(n+r)(n+2r)}} &=\sum_{r=1}^{n}\frac 1{(1+\frac rn)(1+2\frac rn)}\frac 1n\\ &=\int_0^1\frac{\mathrm dx}{(1+x)(1+2x)}\\ &=\log\left(\frac 32\right) \end{align} hence there is a typo in your book.