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

• Jul 21st 2012, 05:49 PM
shaharg
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?
• Jul 21st 2012, 08:01 PM
skeeter
Re: convergency test of ((2n+1)^n)/(n^(2n))
ratio and nth root tests

scroll down to the nth root test section + examples
• Jul 22nd 2012, 08:38 AM
shaharg
Re: convergency test of ((2n+1)^n)/(n^(2n))
Thanx
• Jul 22nd 2012, 08:49 AM
Prove It
Re: convergency test of ((2n+1)^n)/(n^(2n))
The ratio test also works.
• Jul 22nd 2012, 09:00 AM
shaharg
Re: convergency test of ((2n+1)^n)/(n^(2n))
Quote:

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.
• Jul 22nd 2012, 09:42 AM
Plato
Re: convergency test of ((2n+1)^n)/(n^(2n))
Quote:

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.
$\sqrt[n]{{\frac{{{{\left( {2n + 1} \right)}^n}}}{{{n^{2n}}}}}} = \frac{{2n + 1}}{{{n^2}}} \to 0$
• Jul 22nd 2012, 09:46 AM
shaharg
Re: convergency test of ((2n+1)^n)/(n^(2n))
Quote:

Originally Posted by Plato
The root test is really simple.
$\sqrt[n]{{\frac{{{{\left( {2n + 1} \right)}^n}}}{{{n^{2n}}}}}} = \frac{{2n + 1}}{{{n^2}}} \to 0$

Yes it is simple.
I appreciate your effort. Thanx a lot.