How to prove that $1/n!$ is less than $1/n^2$? I want to prove 
$$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test.

How to prove $n^2 < n! $ ?

 A: We can do the comparison with $\dfrac{2}{n^2}$. It is certainly true that $\dfrac{1}{1!}\le \dfrac{2}{1^2}$. 
And for $n \gt 1$, we have $n!\ge (n)(n-1)$. But $n-1\ge \dfrac{n}{2}$, and therefore $n!\ge \dfrac{n^2}{2}$. It follows that $\dfrac{1}{n!}\le \dfrac{2}{n^2}$ for all positive $n$.  
A: Most likely this is proved first time with induction. You  could also take the root test, or see that 
\[ \sum_{n=1}^\infty \frac{1}{n!} = e-1 \]
A: Hint:
$n^2 <n! \implies n(n-(n-1)!) <0$, and note that $(n-1)! >n$ for all $n>3$
A: $2(n-1)>n$
$3(n-2)>n$
$4(n-3)>n$
$\cdots$
$\cdots$
$(k+1)(n-k)>n$
$\cdots$
$\cdots$
$(n-1)2>n$   
Multiplying all, $~$ $(n-1)^{2}\times (n-2)^{2}\times\cdots\times 2^{2}\times 1^{2}>n^{n-2}$   
$\therefore$ $n^{2}\times (n-1)^{2}\times \cdots 2^{2}\times 1^{2}>n^{n}$   
$\therefore$ $n!>n^{\frac{n}{2}}$ $~$ $(n\geq 3)$   
$\therefore$ $n!>n^{\frac{n}{2}}\geq n^{2}$ $~$ $(n\geq 4)$
A: here is a hint. $n! = 1\times2\times...\times(n-2)\times(n-1)\times n$, where as $n^2 = n\times n$. so you have effectively as your step one:
 $1\times2\times...\times (n-1)\times n =n\times n $. 

now you can cancel things out and....well I think that is hint enough as far as the question of proving $n^2 < n!$ goes. 
A: Use induction. Note $k^2<k!$$\implies(k+1)!$$=k!(k+1)$$>k^2(k+1)$$>(k+1)^2.$ That $k^2>k+1$ follows from $k\geq2.$
A: Try the ratio test:
$$
a_n = \frac{1}{n!} \quad \implies \quad \left| \frac{a_{n+1}}{a_n} \right| = \frac{n!}{(n+1)!} = \frac{1}{n+1}
$$
so that
$$
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1,
$$
and thus the series converges absolutely.  Later in your studying series, you may see that
$$
\sum_{n =0}^{\infty} \frac{x^n}{n!} = e^x
$$
which implies
$$
\sum_{n =1}^{\infty} \frac{1}{n!} = \sum_{n = 0}^{\infty} \frac{1^n}{n!} - \frac{1^0}{0!} = e - 1.
$$
