Is $(n + 1)^n = n^n + (n + 1)$? I'm trying to solve my homework, but I don't know if I'm on the right way. 
I have 
$$\frac{(n+1)^n(n+1)}{2(2n+1)n^n}$$
and I want to cancel $(n+1)^n$ with $n^n$.
So I have a sequence of series and try to check the convergence with Ratio Test.
$$\lim_{n\to \infty}\frac{(n+1)^{n+2}}{n^n(2n+1)(2n+2)} $$
 A: Always try with trivial cases at first, notably for $n=1$ we have
$$(n+1)^n=n^n+(n+1) \implies 2=1+2$$

Edit
Since the original problem seems related to the ratio test for convergence, we have that
$$\frac{(n+1)^n(n+1)}{2(2n+1)n^n}=\frac{(n+1)^n}{n^n}\frac{n+1}{4n+2}=\left(1+\frac1n\right)^n\frac{n+1}{4n+2}$$
then take the limit.

Edit 2
For the given limit
$$\lim_{n\to \infty}\frac{(n+1)^{n+2}}{n^n(2n+1)(2n+2)}$$
by ratio test we obtain
$$\frac{(n+2)^{n+3}}{(n+1)^{n+1}(2n+3)(2n+4)}\frac{n^n(2n+1)(2n+2)}{(n+1)^{n+2}}=$$
$$=\frac{n^n(n+2)^{n+3}}{(n+1)^{n+1}(n+1)^{n+2}}\frac{(2n+1)(2n+2)}{(2n+3)(2n+4)}=$$
$$=\frac{n^n(n+2)^{n}}{(n+1)^{n}(n+1)^{n}}\frac{(n+2)^{3}}{(n+1)^{3}}\frac{(2n+1)(2n+2)}{(2n+3)(2n+4)}=$$
$$=\left(\frac{n^2+2n}{n^2+2n+1}\right)^n\frac{(n+2)^{3}}{(n+1)^{3}}\frac{(2n+1)(2n+2)}{(2n+3)(2n+4)}=$$
$$=\left(1-\frac{1}{n^2+2n+1}\right)^n\frac{(n+2)^{3}}{(n+1)^{3}}\frac{(2n+1)(2n+2)}{(2n+3)(2n+4)} \to 1$$
which is not conclusive.
Anyway we don't need ratio test since
$$\frac{(n+1)^{n+2}}{n^n(2n+1)(2n+2)}=\frac{(n+1)^n}{n^n}\frac{(n+1)^{2}}{(2n+1)(2n+2)}=$$
$$=\left(1+\frac1n\right)^n\frac{n^2+2n+1}{4n^2+6n+2} \to e \cdot \frac14 = \frac e 4$$
A: Maybe you need to use the binomial theorem?
$$(n+1)^n=\sum_{j=0}^n \binom{n}{j}n^j1^{n-j}=\sum_{j=0}^n \binom{n}{j}n^j. $$
