Using the Limit Comparison Test on $\sum_{n=1}^{\infty} \frac{n^2} {n!}$

Question

is this right ? $$ \sum_{n=1}^{\infty} \frac{n^2} {n!} $$ i need to use quotient criterion

$$ \lim_{n\to\infty} \frac{\frac{n^2}{n!}}{ \frac{1}{n!}} = \lim_{n\to\infty} {\frac{n^2}{n!}} { \frac{n!}{1}} = \lim_{n\to\infty} {n^2} = \infty $$

so $$ \lim_{n\to\infty} \frac{\frac{n^2}{n!}}{ \frac{1}{n!}} $$ equals $\infty$ and $$\sum_{n=1}^{\infty} \frac{1} {n!} $$ is divergent it means that
$$ \sum_{n=1}^{\infty} \frac{n^2} {n!}$$ is divergent

Answer 1

This doesn't work!

In the Limit Comparison Test, you need $$ \lim_{n\to \infty} \frac{a_n}{b_n} $$ to be a positive real number (and not infinity) when comparing $\sum a_n$ and $\sum b_n$.

Consider the example where $$ a_n = \frac{1}{n} \quad\text{and}\quad b_n = \frac{1}{n^2}. $$ Here $\sum a_n$ is divergent and $\sum b_n$ is convergent, but $$ \lim_{n\to \infty} \frac{a_n}{b_n} = \infty. $$

I don't know how you would use the Limit Comparison Test here. I see now that Landon does know. So you should see his answer!

As a sidenote, note that $$ \sum_{n=1}^{\infty} \frac{1}{n!} $$ is convergent and not divergent.

Answer 2

Can you show that $\dfrac{n^2}{n!}<\dfrac{1}{n^2}$ for sufficiently large $n$?

Then use the monotonic property of the sum from that $n$ onwards. The other $n$'s less than that particular $n$ contribute only to a finite sum.

EDIT: Since the question used the words "Limit Comparison", consider comparing with $\dfrac{1}{(n-2)!}$.

Notice that $$\lim_{n\to\infty}\dfrac{\dfrac{n^2}{n!}}{\dfrac{1}{(n-2)!}}=\lim_{n\to\infty}\dfrac{n}{n-1}=1$$

Note that $\dfrac{1}{n!}<\dfrac{1}{n^2}$ for all $n>3$. Thus $\sum_{n=3}^\infty\dfrac{1}{(n-2)!}$ converges. Hence $\sum_{n=3}^\infty \dfrac{n^2}{n!}$ converges, and hence $\sum_{n=1}^\infty\dfrac{n^2}{n!}$ converges.

Answer 3

Another straightforward use of the limit test can be done in small pieces:

• Show that $n^2\lt 4\cdot(n-1)^2$ for sufficiently large $n$ (hint: this is equivalent to $n\lt 2(n-1)$ for positive $n$; you should be able to solve that inequality algebraically)
• This then gives $\sum_{n=n_0}^\infty \frac{n^2}{n!}\lt 4\sum_{n=n_0}^\infty \frac{(n-1)^2}{n!}$.
• Now, $\frac{(n-1)^2}{n!}=\frac{(n-1)^2}{n\cdot(n-1)\cdot(n-2)!}\lt\frac{(n-1)^2}{(n-1)\cdot(n-1)\cdot(n-2)!}=?$

Can you fill in the rest of the details from here?

(You should also be able to transform this into a more direct use of the limit comparison test proper — the key concept is to shift the series to get rid of that quadratic term up top)

Answer 4

I propose a simpler, direct approach:

\begin{align*} \sum_{n=1}^{\infty}\frac{n^2}{n!}=&\,\sum_{n=1}^{\infty}\frac{n\times n}{n\times(n-1)!}=\sum_{n=1}^{\infty}\frac{n}{(n-1)!}=\sum_{n=1}^{\infty}\frac{n-1+1}{(n-1)!}=\sum_{n=1}^{\infty}\frac{n-1}{(n-1)!}+\sum_{n=1}^{\infty}\frac{1}{(n-1)!}\\ =&\,\sum_{n=0}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}=\underbrace{\frac{0}{0!}}_{=0}+\sum_{n=1}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}+\sum_{n=0}^{\infty}\frac{1}{n!}\\ =&\,\sum_{n=0}^{\infty}\frac{1}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}=2e. \end{align*}