Is this valid? ( (easy?) Limit Manipulation question) I'm current studying series for a calc 2 exam. 
The problem I'm on is to determine the convergence/divergence of this series:
$$\sum_{n=1}^\infty \frac n{(n+1)^n} $$
The ratio test is appropriate here, so I've set it up:
$$\lim_{n\to\infty} \frac {(n+1)}{(n+2)^{(n+1)}} / \frac n{(n+1)^n}$$
which comes out to:
$$\lim_{n\to\infty} \frac {(n+1)^{(n+1)}}{n(n+2)^{(n+1)}}$$
the limit rules say that:
$$\lim_{n\to\infty}(a_n*b_n) = \lim_{n\to\infty}a_n * \lim_{n\to\infty} b_n $$
so
$$\lim_{n\to\infty} \frac {(n+1)^{(n+1)}}{n(n+2)^{(n+1)}} = \lim_{n\to\infty}\frac {(n+1)^{(n+1)}}{(n+2)^{(n+1)}} * \lim_{n\to\infty} \frac 1n$$
I don't really know how to evaluate the first term, (at least not without a bunch of rearranging):
$\lim_{n\to\infty}\frac {(n+1)^{(n+1)}}{(n+2)^{(n+1)}}$, but I do know that $\lim_{n\to\infty} \frac 1n = 0$ , so I get $$\lim_{n\to\infty} \frac {(n+1)^{(n+1)}}{n(n+2)^{(n+1)}} = \lim_{n\to\infty}\frac {(n+1)^{(n+1)}}{(n+2)^{(n+1)}} * 0 = 0$$
On the surface this looks perfectly fine to me, but I have a bit of a weird feeling about it. Is it valid to pull a piece of the limit that you know is equal to 0 out like this? Since anything multiplied by 0 equals 0, is what I've shown here enough to say that:
$$\lim_{n\to\infty} \frac {(n+1)^{(n+1)}}{n(n+2)^{(n+1)}} = 0$$
I realize this may look like a bit of stupid question, but it's been a while since I had to deal extensively with limits, and I want to be sure that my understanding of how to manipulate them is rock solid for the coming exam. Thanks in advance!
 A: Listen to your weird feeling.
The rule $\lim x_ny_n = (\lim x_n)(\lim y_n)$ holds only provided that both limits on the right-hand side exist. So you can't conclude that the limit is $0$ without first knowing that $\lim\frac{(n+1)^{n+1}}{(n+2)^{n+1}}$ converges.
If not for the bolded restriction, you could write
$$ 1 = \lim_{n\to \infty} 1 = \lim_{n\to\infty}n\cdot\frac1n = (\lim_{n\to\infty}n)(\lim_{n\to\infty}\frac 1n)  = (\lim_{n\to\infty}n)0 = 0 $$
which is obviously not true.
A: The only problem with your approach is as follows: (assuming the limit on the left exists) you can only write
$$ \lim_{n\to\infty} \frac {(n+1)^{(n+1)}}{n(n+2)^{(n+1)}} = \lim_{n\to\infty}\frac {(n+1)^{(n+1)}}{(n+2)^{(n+1)}} * \lim_{n\to\infty} \frac 1n $$
if both of the limits on the right exist, and if the product of the limiting values exist.
In particular, if the middle of the three limits doesn't exist, or if it is $+\infty$, then you can't do this step.

There is a variation you can use. You don't actually need the middle limit to exist for the spirit of your argument to work. You could instead talk about limit points, a useful and intuitive concept, but you probably won't be introduced to those in your class. However, something you can work out with what you've leared is the following
Exercise: Prove the theorem

If $-B < f(x) < B$ for all $x$ and $\lim g(x) = 0$, then $\lim f(x) g(x) = 0$.

It's not hard to find a $B$ that works for your particular problem.

That said, you can actually compute the middle limit exactly. And for other problems it is useful to know how to do so. You can actually make it look something like the limit for $e$:
$$ \frac {(n+1)^{(n+1)}}{(n+2)^{(n+1)}}
=
\left(\frac {(n+1)}{(n+2)} \right)^{n+1}
= \left(1 + \frac {1}{(n+2)} \right)^{n+1}
$$
but rather than work through this exercise, it would be easier to use the "logarithm trick". Since we are taking the limit of something nonnegative, we know that
$$ \lim_n \frac {(n+1)^{(n+1)}}{(n+2)^{(n+1)}} = L $$
if and only if
$$ \lim_n \log\left(\frac {(n+1)^{(n+1)}}{(n+2)^{(n+1)}}\right) = \log L $$
(the edge cases here require $\log 0 = -\infty$ and $\log(+\infty) = +\infty$)
The logarithmic limit is easier to compute, since you can use log identities to simplify it, and transform it into another form you know how to deal with.
A: You have
$$ (n +1)^n = n^n (1 + 1/n)^n \sim (ne)^n.$$
The terms of this series decay to zero faster than any exponential, so  the sum converges, and pronto.
A: The $n$-th term is less than $\frac{1}{(n+1)^{n-1}}$, which is less than $\frac{1}{n^2}$ if $n\ge 3$. So by comparison with the convergent series $\sum \frac{1}{n^2}$ our series converges.
Remark: As to the "pulling out," it is no problem.  If both limits on the right exist, then it is true that $\lim_{n\to\infty} a_nb_n=\lim_{n\to\infty} a_n\lim_{n\to\infty} b_n$.
