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?


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.

  • $\begingroup$ Very nice solution, thanks! $\endgroup$ – Mah Jan 7 '15 at 17:03

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