MHB Sum of Infinite Series: Find 1/sqrt(2)

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The discussion centers on finding the sum of an infinite series that equals 1/sqrt(2). The original poster attempted to use the ratio test but found it unhelpful. A response clarifies that the series in question is the power series expansion for cos(π/4). By comparing it to the standard power series for cos(x), the correct result can be derived. This highlights the importance of recognizing series forms in solving such problems.
Umar
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Hey guys! I just have a question regarding finding the sum of an infinite series. Attached is the image of the question. I've tried to use the ratio test but it doesn't give me the result I need which happens to be 1/sqrt(2). I feel like this is one of those power series questions, but I'm not quite sure how to deal with this. If anyone could help out, that would be greatly appreciated!

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Umar said:
Hey guys! I just have a question regarding finding the sum of an infinite series. Attached is the image of the question. I've tried to use the ratio test but it doesn't give me the result I need which happens to be 1/sqrt(2). I feel like this is one of those power series questions, but I'm not quite sure how to deal with this. If anyone could help out, that would be greatly appreciated!
This is just the power series expansion for $\cos\frac\pi4.$
 
This is \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\left(\frac{\pi}{4}\right)^{2n}

Compare that to the power series for cos(x), \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} to get Opalg's result.
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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