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?
 A: Hint: note that the bottom sum is 
$$
\sum_{i=1}^n(2i-1)(2i+1) = \sum_{i=1}^n(4i^2 - 1). 
$$
From here you might be interested in the sum of square integers formula.
A: Method one
We can write the summation formula for both the numerator and denominator which is not my favorite :)
Method two
We use calculus (and the definition of integral) to find the limit.
$$
\lim_{n\to\infty}\frac{1*2+2*3+...+n(n+1)}{1*3+3*5+...+(2n-1)(2n+1)}
{=
\lim_{n\to\infty}\frac{\sum_{i=1}^ni(i+1)}{\sum_{i=1}^n4i^2-1}
\\=
\lim_{n\to\infty}\frac{\sum_{i=1}^ni^2+\sum_{i=1}^ni}{-n+\sum_{i=1}^n4i^2}
\\=
\lim_{n\to\infty}\frac{{1\over n}\sum_{i=1}^n\left({i\over n}\right)^2+{1\over n^2}\sum_{i=1}^n\left({i\over n}\right)}{-{1\over n^2}+{1\over n}\sum_{i=1}^n4\left({i\over n}\right)^2}
\\=
\frac{\int_0^1 x^2dx}{\int_0^1 4x^2dx}
\\={1\over 4}
}
$$
A: 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 $a_n \to \frac{1}{4}$.
A: This is a dirty and rough way of looking at the limit:
$(2n-1)(2n+1)\approx4\cdot n(n+1)$.
Therefore, for large $n$, the denominator is approximately $4$ times the numerator, which means the limit is $\displaystyle \frac{1}{4}$. 
