# Thread: convergency test of ((2n+1)^n)/(n^(2n))

1. ## convergency test of ((2n+1)^n)/(n^(2n))

Hy,
It's suppose to be simple one, but I really dont know how to start and what test should I apply.
((2n+1)^n)/(n^(2n))

Should I work on ((2n+1)/(n^2))^n?

2. ## Re: convergency test of ((2n+1)^n)/(n^(2n))

ratio and nth root tests

scroll down to the nth root test section + examples

Thanx

4. ## Re: convergency test of ((2n+1)^n)/(n^(2n))

The ratio test also works.

5. ## Re: convergency test of ((2n+1)^n)/(n^(2n))

Originally Posted by Prove It
The ratio test also works.
I have tried the ratio test, but putting n+1 instead of every n made it so complicated and stuff didn't cancel to me, I probably do it wrong.

6. ## Re: convergency test of ((2n+1)^n)/(n^(2n))

Originally Posted by shaharg
I have tried the ratio test, but putting n+1 instead of every n made it so complicated and stuff didn't cancel to me, I probably do it wrong.
The root test is really simple.
$\displaystyle \sqrt[n]{{\frac{{{{\left( {2n + 1} \right)}^n}}}{{{n^{2n}}}}}} = \frac{{2n + 1}}{{{n^2}}} \to 0$

7. ## Re: convergency test of ((2n+1)^n)/(n^(2n))

Originally Posted by Plato
The root test is really simple.
$\displaystyle \sqrt[n]{{\frac{{{{\left( {2n + 1} \right)}^n}}}{{{n^{2n}}}}}} = \frac{{2n + 1}}{{{n^2}}} \to 0$
Yes it is simple.
I appreciate your effort. Thanx a lot.