2
$\begingroup$

I want to prove if $\sum\limits_{n=2}^\infty (-1)^n \frac{1}{n!}$ is convergent or not.

I can prove this by the $ratio \ test$.

Isn't this series an alternating series with $a_n = (-1)^n$ and $b_n = \frac{1}{n!}$?

If it is, I can prove with the $Leibniz \ test$ because $b_n>= b_{n+1}$ and $lim_{n\to \infty} b_n = 0$

Is this correct?

$\endgroup$
2
  • $\begingroup$ Yes. You did it correctly. $\endgroup$
    – Mikasa
    Jan 4, 2013 at 19:30
  • $\begingroup$ Yes, series $\sum\limits_{n=2}^\infty (-1)^n \frac{1}{n!}$ converges by Leibniz test. $\endgroup$ Jan 4, 2013 at 19:34

1 Answer 1

6
$\begingroup$

Yes, by taking $b_n=\frac{1}{n!}$ and use the Leibniz test you see that $$\lim_{n\to\infty}b_n=0$$ and $$b_n=\frac{1}{n!}\geq\frac{1}{(n+1)!}=b_{n+1}, n\geq1$$ so the series converges, however using the ratio test we can find more. In fact, you series is absolutely convergent as well($\sum_{0}^{\infty}\frac{1}{n!}=\text{e}$). $$\lim_{n\to\infty}\left|\frac{b_{n+1}}{b_n}\right|=0<1$$ Note that in Leibniz test, you can also take limit of $|b_n|$ instead of $b_n$.

$\endgroup$
3
  • $\begingroup$ Thanks for your explanation $\endgroup$
    – Favolas
    Jan 4, 2013 at 19:52
  • $\begingroup$ @Favolas: I am glad that I could help you. ;-) $\endgroup$
    – Mikasa
    Jan 4, 2013 at 19:53
  • $\begingroup$ Greeeeeeeeaaaat! +1 $\endgroup$
    – amWhy
    Feb 24, 2013 at 0:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .