# Showing that nth root of c_n is equal to nth root of c_n+1 in the limit

#### OhMyMarkov

##### Member
Hello everyone!

I'm trying to show that $\lim \sup \sqrt[n]{c_{n+1}}=\lim \sup \sqrt[n]{c_n}$

This is my attempt:
$\lim \sup \sqrt[n]{c_{n+1}} = \lim \sup \sqrt[m-1]{c_m}=\lim \sup c_m \; ^{\frac{1}{m}}c_m \; ^{\frac{1}{m(m-1)}}$

I'm stuck here, I think I must use some exponential property that says that something decays faster than something or the ratio of two things is zero in the limit...

Any help is appreciated!

Last edited:

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
I'm trying to show that $\lim \sup \sqrt[n]{c_{n+1}}=\lim \sup \sqrt[n]{c_n}{n}$
What if $c_n=1$ for all $n$?

#### OhMyMarkov

##### Member
Ah excuse me LaTeX typo: I meant

I'm trying to show that: $\lim \sup \sqrt[n]{c_{n+1}} = \lim \sup \sqrt[n]{c_{n}}$

I fixed it in the thread