MHB Finding the value at which the series converges

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The discussion focuses on using the Maclaurin series to determine the convergence of the series $$\sum_{n = 0}^{\infty} (-1)^n \frac{\pi^{2n}}{(2n)!}$$. It is noted that this series resembles the Maclaurin series for the cosine function, specifically $\cos(\pi)$, which equals -1. The series converges for all values of $x$, as the Maclaurin series for $\cos(x)$ is valid for any real number. The key takeaway is that the series converges to -1 when evaluated at $\pi$. Overall, the convergence of the series is confirmed through its relation to the cosine function.
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I need to use the maclaurin series to find where this series converges:

$$\sum_{n = 0}^{\infty} (-1)^n \frac{\pi^{2n}}{(2n)!}$$

But I'm not sure how to do this.
 
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Where is $x$? As it is, the series is equivalent to $\cos(\pi)=-1$ and the MacLaurin series for $\cos(x)$ converges for all $x$.
 

Thread 'Proving that convexity implies second order derivative being positive'

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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