Question: Series Convergence for ((-1)^n*n!)/(1*6*11*...*(5n+1))

[harrietstowe]

Homework Statement

Does the series ((-1)^n*n!)/(1*6*11*...*(5n+1)) from n = 0 to [tex]\infty[/tex]
absolutely converge, converge conditionally or diverge?

Homework Equations

The Attempt at a Solution

I did the ratio test for ((-1)^n *n!)/(5n+1)) and I found that it diverges but apparently that is not the correct series to use. I do not understand how to implement the 1*6*11*... part.
Thanks

 

[Char. Limit]

That's an odd series. This is the best way I've been able to write it...

[tex]\sum_{n=0}^{\infty} \frac{\left(-1\right)^n n!}{\prod_{k=0}^{n} 5k+1}[/tex]

I don't know if this helps, but hopefully it does.

EDIT: Seems like an alternating series test would help.

 

[harrietstowe]

That's an odd series. This is the best way I've been able to write it...

[tex]\sum_{n=0}^{\infty} \frac{\left(-1\right)^n n!}{\prod_{k=0}^{n} 5k+1}[/tex]

I don't know if this helps, but hopefully it does.

EDIT: Seems like an alternating series test would help.

What do the columns in the denominator mean? I don't think I have seen that symbol before

 

[Avodyne]

[itex]\prod[/itex] is the product symbol: [itex]\prod_{k=0}^n a_k = a_0 a_1 ... a_n[/itex]

So [itex]n! = \prod_{k=1}^n k[/itex].