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Question

In this exercise, determine the convergence or divergence of the series.

n=0(1)n(2n+1)!\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2 n+1) !}

Solution

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Answered 2 years ago
Answered 2 years ago
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Let n=0(1)nan=n=0(1)n(2n+1)!\sum_{n=0}^{\infty}(-1)^{n}a_n=\sum_{n=0}^{\infty}\dfrac{(-1)^{n}}{(2n+1)!}. Consider

limnan=limn1(2n+1)!=0\begin{aligned}\lim_{n\rightarrow \infty}a_n&=\lim_{n\rightarrow \infty}\dfrac{1}{(2n+1)!}\\&=0\end{aligned}

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