Turning an infinite summation into an integral I was wondering if the following summation can be turned into a Riemann sum ($n\to\infty$):
$$
\sum_{i=1}^{n} \frac{1}{n}\log\left (\frac{a+bi^2}{c+di^2}\right).
$$
Or is there another way to find a closed form answer for this summation?
 A: Assume $a, b, c, d > 0$, we can rewrite the sum as
$$\sum_{k=1}^n \frac1n\log\left(\frac{a+bk^2}{c+dk^2}\right)
= \sum_{k=1}^n \frac1n\left[
\log\frac{b}{d} 
+ \log\left(1 + \frac{a}{bk^2}\right)
- \log\left(1 + \frac{c}{dk^2}\right)
\right]
$$
Notice the last two terms behave like $O(k^{-2})$ for large $k$, 
$$
\sum_{k=1}^\infty \log\left(1 + \frac{a}{bk^2}\right)
\quad\text{ and }\quad
\sum_{k=1}^\infty \log\left(1 + \frac{c}{dk^2}\right)
$$
converge and their contribution to the original sum disappear as $n \to \infty$.
This implies
$$\lim_{n\to\infty} 
\sum_{k=1}^n \frac1n\log\left(\frac{a+bk^2}{c+dk^2}\right) = \log\frac{b}{d}\tag{*1}$$
As an alternative, the sum at hand has the form of a Cesàro mean. It is known that if a sequence $(\alpha_k)$ converges to some limit $L$, then the averages converges to the same limit. i.e.
$$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \alpha_k = L$$
Notice $\displaystyle\;\lim_{k\to\infty} \log\left(\frac{a+bk^2}{c+dk^2}\right) = \log\frac{b}{d}$, $(*1)$ follows immediately.
