Is there a good estimate for $\sum_n\frac{1}{\cos(n\theta)}$?

Question

Is there a good estimate for the following series (when $\theta$ is very small)?

$$\sum_{n=1}^{\lfloor\frac{\pi}{4\theta}\rfloor}\frac{1}{\cos(n\theta)}$$

My original problem is:

Given a right trangle $OAB$ where A is the right angle; additionally we assume that $OA=AB$. Now we divide AB into $m$ equal parts by $m+1$ points (including $A$ and $B$.) Then connect $O$ with each of these $m+1$ points. Estimate the sum of these $m+1$ segments.

I know my problem is not the same as the series above, so any idea to estimate the series or to attack my original problem is welcome.

Exact or rigorous solutions are unnecessary for me because this is actually an engineering problem. Some approach which "kinda makes sense" is good enough for me.

Answer 1

The following addresses the original question that was asked by the OP.

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One could approximate the sum of the diagonal segments by using Riemann sums.

The sum of the lengths of the diagonal segments, $S(m)$, is given by

$$S(m)=m\times OA\left(\frac1m \sum_{n=1}^m \sqrt{1+(n/m)^2}\right)$$

The parenthetical term is the Riemann sum for

$$\begin{align}\int_0^1\sqrt{1+x^2}\,dx &=\frac12(\sqrt 2 +\text{sinh}^{-1}(1))\\\\ &\approx 1.1478 \end{align}$$

So, for large $m$, the sum can be approimated by

$$\begin{align} S(m)&\approx \frac12(\sqrt 2+\text{sinh}^{-1}(1)) \,(m\times OA)\\\\ & \approx 1.1478\,( m\times OA) \end{align}$$