Calculating $\lim_{n \to \infty}\frac{1*2+2*3+...+n(n+1)}{1*3+3*5+...+(2n-1)(2n+1)} $ This might seem pretty straightforward, as both the numerator and the denominator seem to resemble some well-known sums, however the numerator is the only one which can actually be rewritten as $\frac{n(n+1)(n+2)}{3}$. Could you point me to the right direction as to how I'd be able to continue the exercise in this manner, or is there perhaps a different way of dealing with this limit?

Let's assign the series as $a_n$. Notice, that:
$$b_n = \frac{\sum_{i=1}^n i (i+1)}{\sum_{i=1}^n (2i - 2) 2i}=
\frac{1}{4} \frac{\sum_{i=1}^n i (i+1)}{\sum_{i=1}^n (i - 1) 2i}<a_n<
\frac{\sum_{i=1}^n i (i+1)}{\sum_{i=1}^n 2i \cdot 2i} = 
\frac{1}{4}\frac{\sum_{i=1}^n i (i+1)}{\sum_{i=1}^n i^2} = c_n$$
Also $b_n \to \frac{1}{4}$ and $c_n \to \frac{1}{4}$. Using squeze theorem we get $a_n \to \frac{1}{4}$.
