Which test would be appropriate to use on this series to show convergence/divergence? $$\sum_{n=1}^\infty \frac{(2n+1)^n}{n^{2n}}$$
I am struggling to come up with something. First thought root test but I don't think that will work since the roots aren't the same? I'm guessing maybe the ratio test will work since it usually does but is there a more efficient test to use here?
 A: Hint: for any $n \geq 6$,
$$
\frac{{(2n + 1)^n }}{{n^{2n} }} \le \frac{{(3n)^n }}{{(n^2 )^n }} = \bigg(\frac{3}{n}\bigg)^n  \le \bigg(\frac{1}{2}\bigg)^n .
$$
A: HINT:
Both the ratio test (with a few generous inequalities) and the root test (definitely the best bet) work here.
Also, there is a general order that I think of and that I told my students when they tested for convergence. It is by no means infallible. Here's what I would say, and in this order:


*

*Write out a few terms! Get a feel for the series.

*Make sure the limit of the terms goes to zero. 

*If it's integrable, use the integral test.

*If it alternates/telescopes, try the appropriate alternating/telescoping series tests.

*If it has a factorial, use the ratio test.

*If it has things raised to nth powers (like this one), use the root test.

*Use comparison - whichever feels more natural (I think basic is easier to see than limit, but so it goes)


I repeat, this is not infallible. Comparison tests can be scary, and some series are brilliantly bounded using astounding combinations of tests and ingenuity.
A: The root test should work, since
$$
\sqrt[n]{\frac{(2n+1)^n}{n^{2n}}}=\frac{2n+1}{n^2}.
$$
It doesn't matter that the exponents of the numerator and denominator are not quite the same.
A: You can rewrite it so that the powers are the same: $n^{2n} = (n^2)^n$. Then use the root test.
