# Using a comparison test for series with factorials and repeating patterns

I am doing a homework problem where we need to use the appropriate comparison test (Direct or limit) to determine if the following series is convergent of divergent:

$$\sum_{i=1}^{\infty} \frac{i!}{5\cdot8\cdot11\cdot\cdot\cdot(3i+2)}$$

Of course, the ratio test would be straight forward for this case. However we have not yet gotten to it. The question specifically states that we must use a comparison test.

My first thought is that this is convergent. The denominator is getting larger than the numerator faster. I would imagine doing a direct comparison test would be the best. I would just need to find a value that is greater than i! but less than the denominator that still results in either geometric series or p series.

Am I taking the wrong approach?

Here is a good way to use the comparison test:

\begin{align}5\times8\times11\times\dots\times(3i+2)&>2\times4\times6\times\dots\times(2i)\\&=(2\times1)(2\times2)(2\times3)\dots(2\times i)\\&=2^ii!\end{align}

Thus,

$$\frac{i!}{5\cdot8\cdot11\cdot\cdot\cdot(3i+2)}<\frac{i!}{2^ii!}=\frac1{2^i}$$

And this is just a geometric series that converges.

• This is perfect! Is there any way to easily identify these types of patterns? Am I almost always trying to get a form of 2^i? Mar 9, 2017 at 22:25
• No, you try to look for some sort of thing such that you end up with $a^n i!$, at least here, since you want the factorial to cancel off from the numerator. Now, one could've easily done $3^i$ instead, and I'll leave that for you to show (it's very easy and direct like my answer) Mar 9, 2017 at 22:27
• In your solution, you got 5 x 8 x 11 x ... x (3i+2) < 2 x 4 x 6 ... (2i). Once you switch that to the denominator though, shouldn't the equality also flip around? Mar 9, 2017 at 22:40
• Sorry, my first inequality was backwards. Mar 9, 2017 at 22:41
• Ah, thank you for your help! Mar 9, 2017 at 22:41