# Convergence or divergence of series involving $(-1)^n$

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?

• Are you trying to say something like the alternating series test? Nov 3, 2021 at 22:27
• Do you know the power series for the exponential function?
– user987907
Nov 3, 2021 at 22:27
• 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 $\lim{n\to\infty}\frac{(-1)^n}{2n-1}$? Nov 3, 2021 at 22:29
• @JoséCarlosSantos The convergence of the series, if they do in fact converge. Nov 3, 2021 at 22:36
• @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.) Nov 3, 2021 at 22:41

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 $$\sum_{n=0}^\infty (-1)^n\frac{1}{n!}=e^{-1}\;.$$