Proving convergence or divergence using the comparison test

Solution 1:

Since $n!$ grows faster than $n^{300}$, there will be some value $N$ at which we have the inequality $n!>n^{300}$ for all $n>N$. You can use this inequality in a comparison:

$$\sum_{n=1}^{\infty}\frac{n!}{n^{300}}=\sum_{n=1}^N\frac{n!}{n^{300}}+\sum_{n=N+1}^{\infty}\frac{n!}{n^{300}}>\sum_{n=1}^N\frac{n!}{n^{300}}+\sum_{n=N+1}^{\infty}1$$

See if you can complete the reasoning to show that the original series diverges.

To see that $n!$ grows faster than $n^k$ for any positive integer $k$, notice for $n>k$ we have:

$$\frac{n!}{n^k}=(n-k)!\cdot\frac{n-k+1}{n}\cdot\frac{n-k+2}{n}\cdot\ldots\cdot\frac{n-1}{n}\cdot\frac{n}{n}$$

Now, choose $n$ big enough so that $2(n-k+1)>n$. It follows that each of the $k$ fractions above are greater than $1/2$. This gives us the inequality:

$$\frac{n!}{n^k}>(n-k)!\cdot\left(\frac{1}{2}\right)^{k}$$

Remembering that $k$ is fixed (in our case, $k=300$), as we let $n$ grow, eventually the right hand side of the above inequality will be greater than $1$, so that $n!>n^k$.

Solution 2:

Hint: Check the limit of the summand first

$$ \lim_{n\to \infty} \frac{n!}{n^{300}}=\infty. $$

Now, since the limit is $\infty$, then series diverges.

Note:

1) If a series $\sum b_n $ converges, then $\lim_{n\to \infty} b_n = 0$, which implies if $\lim_{n\to \infty} b_n \neq 0 $, then the series diverges.

2) You can use the following result to find the above limit

If $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}=a $ and $|a|<1$, then $\lim_{n \to \infty} a_n =0 $.