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,
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
Which limit would you say is more obvious?
$\displaystyle \lim_{n\to \infty}\frac{\frac{(10)^{n+1}}{(n+1)!}}{\frac{(10) ^n}{n!}}=\lim_{n\to \infty}\frac{10}{n+1}=?$
$\displaystyle \lim_{n\to \infty}\left(\frac{(10)^n}{n!}\right)^{1/n}=\lim_{n\to \infty}\frac{10}{(n!)^{1/n}}=\lim_{n\to \infty}\frac{10}{e^{\frac{\ln(n!)}{n}}}=?$
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!}} $.
Look at this page.
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?