# Sum this series: $1+\cos{\theta}\sec{\theta}+\cos{2\theta}\sec^2{\theta}+\cdots+\cos{n\theta}\sec^n{\theta}$

Sum this series: $$1+\cos{\theta}\sec{\theta}+\cos{2\theta}\sec^2{\theta}+\cdots+\cos{n\theta}\sec^n{\theta}$$ This problem comes from the Sixth Term Examination Paper III 1990.

I have proved it using induction, but the previous part of the question derived two identities: $$\cos{\alpha}+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+2n\beta)=\frac{\sin(n+1)\beta\cos(\alpha+n\beta)}{\sin{\beta}}$$ And $$\cos{\alpha}+\binom{n}{1}\cos(\alpha+2\beta)+\cdots+\binom{n}{r}\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta)=2^n \cos^n \beta \cos ( \alpha + n \beta)$$ I am stuck to establish a relation between these two identities and the sum of the series.

• It's the real part of $\sum e^{ik\theta}\sec^k\theta$. – Lord Shark the Unknown Apr 3 at 4:14
• Welcome to Math.SE! Please always try to include something of what you know about a problem or where exactly you got stuck. Isolated problem statements tend to attract down- and close-votes, but more importantly: Information you provide helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) telling you things you already know or using techniques you aren't yet expected to know. (Plus, it helps convince people that you aren't simply trying to get them to do your homework for them.) – Blue Apr 3 at 4:22
• pastpapercache.blob.core.windows.net/ppppapers/step/… – Tony Tong Apr 3 at 4:52
• It is the fourth question. – Tony Tong Apr 3 at 4:54
• @TonyTong: Thanks for the clarification! :) I believe you'll agree that establishing a relationship between the identities and the sum is quite a different problem than merely summing the series by whatever means. (Thus HoldingArthur's detailed response may be rendered irrelevant.) The fact that you've already proven the result is also helpful; usually, I would recommend including the proof in your question (so that answerers don't waste (more) time duplicating your effort), but, given the nature of your actual question, that proof also seems irrelevant. – Blue Apr 3 at 5:51

If you remember the complex numbers, you know that $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ We can sum the following series first by formulae of geometric sequences: $$\sum_{k=0}^n e^{ik\theta} \sec^k{\theta}=\sum_{k=0}^n (e^{i\theta} \sec{\theta})^k=\frac{1-(e^{i\theta} \sec{\theta})^{n+1}}{1-e^{i\theta} \sec{\theta}}$$ Similarly, $$\sum_{k=0}^n e^{-ik\theta} \sec^k{\theta}=\sum_{k=0}^n (e^{-i\theta} \sec{\theta})^k=\frac{1-(e^{-i\theta} \sec{\theta})^{n+1}}{1-e^{-i\theta} \sec{\theta}}$$ Therefore, $$\sum_{k=0}^n \cos(k\theta)\sec^k(\theta)=\frac{1}{2}\left(\sum_{k=0}^n e^{-ik\theta}\sec^k{\theta}+\sum_{k=0}^n e^{ik\theta}\sec^k{\theta}\right)\\ =\frac{1}{2}\left(\frac{1-(e^{-i\theta} \sec{\theta})^{n+1}}{1-e^{-i\theta} \sec{\theta}}+\frac{1-(e^{i\theta} \sec{\theta})^{n+1}}{1-e^{i\theta} \sec{\theta}} \right)\\ =\frac{1}{2}\left(\frac{2-2\cos\theta\sec\theta+2\cos(n\theta)\sec^{n+2}\theta-2\cos((n+1)\theta)\sec^{n+1}\theta}{1-e^{i\theta}\sec\theta-e^{-i\theta}\sec\theta+\sec^2\theta} \right)\\ =\csc\theta\sec^n\theta\sin[(n+1)\theta]$$ I will let you to fil in the missing details.