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Prove $\sin(\pi/2)=1$ using the Taylor series definition of $\sin x$, $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

It seems rather messy to substitute in $\pi/2$ for $x$. So we have $$\sin(\pi/2)=\sum_{n=0}^{\infty} \frac{(-1)^n(\pi/2)^{2n+1}}{(2n+1)!}.$$

I'm not too sure where to go from here. Any help would be appreciated! Thanks!

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    @Caligirl11: Please try to learn some mark-up for posting your questions, rather than relying on others to make them readable. Also: you have posted five other questions, and have yet to accept any answer. If a prior question has been answered to your satisfaction, be sure to formally "accept" an answer (the on you find most helpful, best, etc) by clicking on the checkmark next to the question.2011-04-19
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    Is this a homework question? It's very strange.2011-04-19
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    Ok sure thank you. I'm sorry I'm new to this website. I wasn't aware of the checkmark system. I will do that from now on.2011-04-19
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    Doesn't seem strange to me. Someone comes up to you with an infinite series. If you don't recognize it as the Taylor series for $\sin x$, evaluated at $x=\pi/2$, can you still contrive to prove it converges to $1$?2011-04-19
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    @Gerry: I mean it's strange in the sense that most calculus textbooks I can think of would not teach the kind of thinking necessary to solve a problem like this (unless there's an easy solution I'm missing).2011-04-19
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    Mosquito-nuker's solution: let $f(t)=\frac1{1+t^2}$ and $g(t)=\int_{-1}^t f(u)\mathrm du$; prove $g(1)=\frac{\pi}{2}$, compose the series for $\sin\;t$ and $g(t)$ ... you get the drift.2011-04-19
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    @Qiaochu, yes, it would be strange to see this problem in a Calculus textbook (except, maybe, Apostol). I assumed it just came to caligirl whilst she was contemplating the sine function.2011-04-19
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    Yes so basically the problem is to show that the infinite series converges to 1. It is easy to show that it converges by the ratio test, I'm just not sure how to show it converges to 1.2011-04-19
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    How are you defining $\pi$ here? What facts about $\sin(x)$ are you allowed to use?2011-04-19
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    @hawaii99 - If you don't mind taking the time to choose an acceptable answer (by clicking the "check mark" below the number of votes for that question), that would be appreciated.2011-11-09

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