how to determine if this series converges? I was trying to find out if the series: 
$$\sum^{\infty}_{n=1}n^3e^{-n} $$
converges. 
I tried applying the Cauchy test, 
$\displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{n^3e^{-n}}=\displaystyle \lim_{n \rightarrow \infty}n^{\frac{3}{n}}e^{-1}$
but that gives an undetermined form of $[ \infty^{0}] $ which doesn't allow me to continue....
any ideas? 
 A: Use a ratio test :$$\lim _{ n\rightarrow \infty  }{ \left| \frac { { a }_{ n+1 } }{ { a }_{ n } }  \right| = } \lim _{ n\rightarrow \infty  }{ \left| \frac { { \left( n+1 \right)  }^{ 3 }{ e }^{ -\left( n+1 \right)  } }{ { n }^{ 3 }{ e }^{ -n } }  \right| =\lim _{ n\rightarrow \infty  }{ \left| \frac { { \left( n+1 \right)  }^{ 3 } }{ e{ n }^{ 3 } }  \right|  }  } =\frac { 1 }{ e } <1$$
so it converges
A: Your approach can be continued. As mentioned in the comments $n^{3/n} \to 1$ as $n \to \infty$. Note that this follows if we can prove $n^{1/n} \to 1$ as $n \to \infty$. Here is an elementary proof using only the (generalized) squeeze theorem and basic algebra.
Take any real $ε > 0$, and let $c \in \mathbb{N}$ such that $1/2^c \le ε$.   [This arises from what we need below.]
As $n \to \infty$:
  Let $k \in \mathbb{N}$ such that $2^k \le n \le 2^{k+1}$.
  Then $k \to \infty$   [since $2^k \ge n/2 \to \infty$].
  Let $d = k-c$.
  Thus $n^{1/n} \le (2^{k+1})^{1/2^k} \le 2^{(2d+1)/2^{c+d}} \le 2^{1/2^c}$ (eventually)   [since $c \le d$ and $2d+1 \le 2^d$].
  Also $2^{1/2^c} \le 1+1/2^c \le 1+ε$.
  Thus $1 \le n^{1/n} \le 1+ε$.
Therefore by the squeeze theorem $n^{1/n} \to 1$ as $n \to \infty$.
A: Hint


*

*$\lim n^k e^{-n} = 0$ for all $k$

*$\sum \frac{1}{n^2}$ is convergent

A: This is a positive series. Let $u_n = n^3 e^{-n}$ such that the given series is $\sum u_n$. One has :
$$ \frac{u_{n+1}}{u_n} = \frac{(n+1)^3}{e^{n+1}} \times \frac{e^n}{n^3}  \xrightarrow[n \longrightarrow \infty]{} \frac{1}{e}$$
By the ratio test, the series $\sum n^3 e^{-n}$ converges.
A: Hint: This series can be summed exactly.
Write $$f(x)=\sum_{n=0}^{\infty}x^n = (1-x)^{-1}$$ which converges absolutely for $|x|<1$. Note that your series can be written in terms of various derivatives $f^{(k)}(e^{-1})$, all of which converge similarly.
Details left to the reader.
