Determine whether the infinite series $\sum_{n=1}^{\infty}\frac{n!}{n^3}$ is convergent

The question asks to use the direct comparison test to determine whether

$$\sum_{n=1}^{\infty}\frac{n!}{n^3}$$

is convergent or divergent.

I was wondering whether the direct comparison test requires that a series consist of a positive sequence, as the only thing I could think of to compare the series to was:

$$\sum_{n=1}^{\infty}-\frac{n^2}{n^3} \leq \sum_{n=1}^{\infty}\frac{n!}{n^3}$$

with the LHS being a negative harmonic series that diverges and hence shows the series on the right diverges as well.

Is this reasoning correct/is there an easier way to do this?

• No, the reasoning is not at all correct: The series whose terms are all zero satisfies the same inequality. – user296602 Mar 7 '16 at 2:07
• Okay, so what do you suggest I do instead? @T.Bongers – patrickh Mar 7 '16 at 2:08
• Why don't you just start by computing, say, the first $10$ terms of the sequence? – user296602 Mar 7 '16 at 2:09
• I could do that, but the question asks specifically to use the direct comparison test. I am looking for a series to compare it to, I already know it diverges. – patrickh Mar 7 '16 at 2:09
• Well, if you compute each term and notice they're all larger than $1$ you should be able to complete the problem. – user296602 Mar 7 '16 at 2:11

Notice that $n! > n^3$ for all sufficiently large $n$, and

$$\sum_{n = 1}^{\infty} 1 = \infty$$

Hint: Use the ratio test. We may drop the absolute values because the terms are always greater than zero.

$$\lim\limits_{n \to \infty}\left|\frac{(n+1)!}{(n+1)^3}\cdot\frac{n^3}{n!}\right|=\lim\limits_{n \to \infty}\frac{n^3(n+1)}{(n+1)^3}=\lim\limits_{n \to \infty}\frac{n^3}{(n+1)^2}$$

Which obviously goes to infinity. Thus, the series diverges. Thanks to T. Bongers for helping.

• Why on earth would you use L'Hospital's rule to compute that limit? – user296602 Mar 7 '16 at 2:11
• Yes, it works; but it's unnecessarily difficult and obfuscates what's actually going on.... – user296602 Mar 7 '16 at 2:12
• Notice that it's equal to $$(n + 1) \cdot \frac{1}{(1 + 1/n)^3}$$ as an application of the standard technique of dividing by the highest-order term. – user296602 Mar 7 '16 at 2:13
• @zz20s $n^3/(n+1)^3\to 1$ clearly, so the limit you gave clearly goes to infinity. – Tim Raczkowski Mar 7 '16 at 2:14