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Thoughts about the convergence (conditional or absolute) and divergence of series:

I'm trying to prove that these series converge conditionally, converge absolutely, or diverge.

$\sum^∞_\mathrm{n=0}(−1)^n\frac{1}{n!}$

My initial thought was that the sequence must converge to $0$. The sign of each term in the sequence of partial sums will depend entirely on $(-1)^n$ as $n \geq 0$.

We know $0!$ is $1$, hence the sequence of partial sums has $1$ as its first term. Each subsequent term will be smaller than the last, because $\frac{1}{n!} > \frac{1}{(n+1)!}$. I see the overall sequence converging as two subsequences approaching $0$ from above and below [$n=1$ gives $(-1)*1$, $n=2$ gives $(-1)^2*\frac{1}{2*1} = \frac{1}{2}$].

Say you had another series. This time:

$\sum^∞_\mathrm{n=0}(−1)^n\frac{1}{(2n-1)}$

By the logic for the first series, this series would also converge. Momentarily ignoring the sign given by $(-1)^n$, each term $\frac{1}{(2n-1)}$ would be greater than the subsequent term $\frac{1}{(2(n+1) - 1)}$. Once you apply the $(-1)^n$, the sequence would also approach 0 from above and below.

I don't know how to wrap these thoughts up into a more formal proof. Am I looking at this the right way?

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    $\begingroup$ Are you trying to say something like the alternating series test? $\endgroup$
    – WhatsUp
    Nov 3, 2021 at 22:27
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    $\begingroup$ Do you know the power series for the exponential function? $\endgroup$
    – user987907
    Nov 3, 2021 at 22:27
  • $\begingroup$ Are you studying the convergence of the series $\sum_{n=0}^\infty\frac{(-1)^n}{n!}$ and $\sum_{n=0}^\infty\frac{(-1)^n}{2n-1}$ or the limits $\lim_{n\to\infty}\frac{(-1)^n}{n!}$ and $\lim{n\to\infty}\frac{(-1)^n}{2n-1}$? $\endgroup$ Nov 3, 2021 at 22:29
  • $\begingroup$ @JoséCarlosSantos The convergence of the series, if they do in fact converge. $\endgroup$ Nov 3, 2021 at 22:36
  • $\begingroup$ @WhatsUp The alternating series test seems to provide a direct approach to this. Thank you for pointing that out. The monotonic condition follows naturally, but proving that limit approaches $0$ is a bit more involved. (I assume you could use the delta-epsilon definition of limit.) $\endgroup$ Nov 3, 2021 at 22:41

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You are thinking too hard. This series is absolutely convergent: simply apply the ratio test to $\sum \frac{1}{n!}$.


If you know the power series for the exponential function, then you can see that $$ \sum_{n=0}^\infty (-1)^n\frac{1}{n!}=e^{-1}\;. $$

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