# Thread: Ratio and Root Test

1. ## Ratio and Root Test

Hello everyone,

I have this problem:$\displaystyle \sum_{n = 0}^\infty (-10)^n/n!$ and I am trying to do the root test, what is stopping me is that I don't know what to do with factorial when I take the soot of the summation.

Thanks,

Hello everyone,

I have this problem:$\displaystyle \sum_{n = 0}^\infty (-10)^n/n!$ and I am trying to do the root test, what is stopping me is that I don't know what to do with factorial when I take the soot of the summation.

Thanks,
Do you have to use the root test? The ratio test is much easier to use here.

3. No, I can use either the ratio or root test, I just assumed that the root test would be easier. Why is the ratio test easier?

Thanks

4. Which limit would you say is more obvious?

$\displaystyle \lim_{n\to \infty}\frac{\frac{(10)^{n+1}}{(n+1)!}}{\frac{(10) ^n}{n!}}=\lim_{n\to \infty}\frac{10}{n+1}=?$
$\displaystyle \lim_{n\to \infty}\left(\frac{(10)^n}{n!}\right)^{1/n}=\lim_{n\to \infty}\frac{10}{(n!)^{1/n}}=\lim_{n\to \infty}\frac{10}{e^{\frac{\ln(n!)}{n}}}=?$

5. I see what you're getting at, I just don't understand how you got rid of the n! using the ratio method.

6. $\displaystyle \frac{(n+1)!}{n!}=\frac{(n+1)\cdot n \cdot (n-1) \cdot ... \cdot 1}{n\cdot (n-1) \cdot ... \cdot 1}=n+1$

I have this problem:$\displaystyle \sum_{n = 0}^\infty (-10)^n/n!$ and I am trying to do the root test.
Please note that you cannot use either the ratio test nor the root test here.
Both of those test are for positive series.
This series is alternating.
$\displaystyle \sum\limits_{n = 0}^\infty {\left( { - 1} \right)^n \frac{10^n }{n!}}$.

8. Haha, that's quite funny because that's a practice problem my prof gave for the Ratio/Root test. I never learn't that these were for positive functions only. I did learn that you take the absolute value of the series, so wouldn't it work in that case because the series isn't actually negative?

If you do have the absolute value of each term then $\displaystyle \sum\limits_{n = 0}^\infty {\frac{{10^n }}{{n!}}}$ then you can use the root test.
$\displaystyle \sum\limits_{n = 0}^\infty {\left( { - 1} \right)^n \frac{10^n }{n!}}$.