Proving convergence by the ratio test. I try to prove by the ratio test that this serie converge :
$$\sum_{n=1}^{\infty} \frac{(2n+1)^n}{n^{2n}}$$
I know I have to demonstrate that
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$$
But, I cannot seem to be able to simplify the following expression,
$$\left| \frac{(2(n+1)+1)^{n+1}}{(n+1)^{2(n+1)}} \cdot \frac{n^{2n}}{(2n+1)^n} \right| $$
Can you give me a hint!?
 A: The Root Test is in this case more pleasant for proving convergence. However, if Ratio Test it must be, rewrite the expression for the ratio as
$$\frac{2n+3}{(n+1)^2} \cdot \left(\frac{2n+3}{2n+1}\right)^n\cdot \left(\frac{n}{n+1}\right)^{2n}.$$
The first term above has limit $0$. Let us now look at the third term. Its reciprocal is $\left(1+\frac{1}{n}\right)^{2n}$, which has limit $e^2$.
Similarly, we can handle the slightly messier second term, which is $\left(1+\frac{2}{2n+1}\right)^n$. A little manipulation, rewriting the exponent as $(2n+1)(n/(2n+1)$, shows this has limit $e$.
So the first term has limit $0$, and the other two terms are bounded above. Thus our $\frac{a_{n+1}}{a_n}$ has limit $0\lt 1$, and therefore we have convergence.
Remark: We could have computed less. For example the third term is clearly less than $1$, so we don't need to worry about it. And we don't need to find the exact limit of the second term, as long as we show it is bounded above.
A: $$\left| \frac{(2n+3)^{n+1}}{(n+1)^{2n+2}} \cdot \frac{n^{2n}}{(2n+1)^n} \right|<\left| \frac{(2n+3)^{n+1}}{n^{2n+2}} \cdot \frac{n^{2n}}{(2n+1)^n} \right| =\frac{(2n+3)^{n+1}}{n^2*(2n+1)^n}=\frac{1}{n^2}*\frac{(2+\frac{3}{n})^{n+1}*n}{(2+\frac{1}{n})^n}<\frac{2n}{n^2}=\frac{2}{n}$$Since $$\lim_{n \to \infty} \left| \frac{2}{n} \right|=0,$$ We have $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|<1$$
Thus, the series converges
