- Thread starter
- #1
Chipset3600
Member
- Feb 14, 2012
- 79
I found this on the internet, but did not find the proof.
Curious to me is that the the ratio and root test have the same conditions.
How can i basically prove this equality?
\(\displaystyle \frac{a_{n+1}}{a_{n}} = \sqrt[n]{a_{n}}\)
Thank you!
Curious to me is that the the ratio and root test have the same conditions.
How can i basically prove this equality?
\(\displaystyle \frac{a_{n+1}}{a_{n}} = \sqrt[n]{a_{n}}\)
Thank you!