What test can I use on the following series in order to determine(adsbygoogle = window.adsbygoogle || []).push({});

if it converges or diverges? Looking at it graphically it appears to

diverge but I cannot show it analytically.

[tex]

\sum\limits_{n = 1}^\infty {\frac{{n!}}{{3n! - 1}}}

[/tex]

Using the Ratio Test, here is I got thus far

[tex]

\left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \left| {\frac{{\left( {\frac{{(n + 1)!}}{{3(n + 1)! - 1}}} \right)}}{{\left( {\frac{{n!}}{{3n! - 1}}} \right)}}} \right|

[/tex]

[tex]

= \frac{{(n + 1)!}}{{3(n + 1)! - 1}} \cdot \frac{{3n! - 1}}{{n!}}

[/tex]

[tex]

= \frac{{(n + 1)n!}}{{3(n + 1)n! - 1}} \cdot \frac{{3n! - 1}}{{n!}}

[/tex]

[tex]

= \frac{{(n + 1)(3n! - 1)}}{{3(n + 1)n! - 1}}

[/tex]

At this point, I don't see how the Ratio test will work. What test can I use on this series?

Should I start off with long division to reduce the improper rational

expression?

http://img164.imageshack.us/img164/4656/longdiv7nr.jpg [Broken]

So the equivalent sum would be

[tex]

\sum\limits_{n = 1}^\infty {\left( {\frac{1}{3} + \frac{1}{{3(3n! - 1)}}} \right)}

[/tex]

Which must diverge because of the 1/3 always being summed from

1 to infinity.

Is that the correct way to do it? If so, is it the only way? Am I

missing a simple test that could be used?

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# Homework Help: What convergence/divergence test can I use on this series? - Calculus II

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