# Calculate lim n->inf (n^2/n!)

1. Apr 20, 2015

### cloveryeah

1. The problem statement, all variables and given/known data
calculate lim n->inf (n^2/n!)

2. Relevant equations

3. The attempt at a solution
i just calculated, and i got inf as the ans...
n^2/1X2X3...Xn
n/1X2X3...X(n-1)
1/[(1X2X3...)/n](1-1/n)
when n->inf, it will go to inf

am i correct? or there is something wrong?

2. Apr 20, 2015

### SammyS

Staff Emeritus
That is incorrect.

3. Apr 20, 2015

### cloveryeah

what is wrong?

4. Apr 20, 2015

### Staff: Mentor

The 2nd and 3rd lines above are OK but written ambiguously.
According to the standard rules of precedence, n/1X2X3...X(n-1) = $\frac n 1 2 \cdot 3 \cdot \dots (n - 1)$, which is probably not what you intended.
The 3rd line would be interpreted as $\frac 1 { \frac{1 \cdot 2 \cdot 3 \dots } n} (1 - 1/n)$. That's so convoluted that you probably misled yourself.

5. Apr 20, 2015

### cloveryeah

why not
n / [1X2X3...(n-1)]

6. Apr 20, 2015

### Staff: Mentor

That would be fine, although at the calculus level, you should get out of the habit of writing X for multiplication. It's too easily confused as a variable.

So what is $\lim_{n \to \infty}\frac n {1 \cdot 2 \cdot 3 \cdot \dots (n - 1)}$?

7. Apr 20, 2015

### cloveryeah

lim→∞ 1/ [ (1⋅2⋅3⋅⋅⋅/n)(1-1/n)] = 1/0 = inf ??

8. Apr 20, 2015

### Ray Vickson

You tell us.

9. Apr 20, 2015

### Staff: Mentor

$\frac {3^2} {3!} = \frac 9 6 = 3/2$
$\frac {4^2} {4!} = \frac {16}{24} = 2/3$
$\frac {5^2} {5!} = ?$
$\frac {6^2} {6!} = ?$
Do you see any sort of trend here?

10. Apr 20, 2015

### pasmith

Consider: If $n > 4$ then $n! = n(n-1) \times \cdots \times 1 > n(n-1)$. Thus $0 < \frac{n^2}{n!} < \frac{n^2}{n(n-1)} < \frac43$. That tells you that the limit of $\frac{n^2}{n!}$, if it exists, must be finite. Unfortunately this bound on $\frac1{n!}$ is not sufficiently accurate for the squeeze theorem to assist, since $\lim_{n \to \infty} \frac{n}{n-1} = 1$.

But how, for $n > 5$, could you modify the above argument so that the squeeze theorem will assist?