- #1
opticaltempest
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What test can I use on the following series in order to determine
if it converges or diverges? Looking at it graphically it appears to
diverge but I cannot show it analytically.
<br /> \sum\limits_{n = 1}^\infty {\frac{{n!}}{{3n! - 1}}}<br />
Using the Ratio Test, here is I got thus far
<br /> \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|<br />
<br /> = \frac{{(n + 1)!}}{{3(n + 1)! - 1}} \cdot \frac{{3n! - 1}}{{n!}}<br />
<br /> = \frac{{(n + 1)n!}}{{3(n + 1)n! - 1}} \cdot \frac{{3n! - 1}}{{n!}}<br />
<br /> = \frac{{(n + 1)(3n! - 1)}}{{3(n + 1)n! - 1}}<br />
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
So the equivalent sum would be
<br /> \sum\limits_{n = 1}^\infty {\left( {\frac{1}{3} + \frac{1}{{3(3n! - 1)}}} \right)} <br />
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?
if it converges or diverges? Looking at it graphically it appears to
diverge but I cannot show it analytically.
<br /> \sum\limits_{n = 1}^\infty {\frac{{n!}}{{3n! - 1}}}<br />
Using the Ratio Test, here is I got thus far
<br /> \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|<br />
<br /> = \frac{{(n + 1)!}}{{3(n + 1)! - 1}} \cdot \frac{{3n! - 1}}{{n!}}<br />
<br /> = \frac{{(n + 1)n!}}{{3(n + 1)n! - 1}} \cdot \frac{{3n! - 1}}{{n!}}<br />
<br /> = \frac{{(n + 1)(3n! - 1)}}{{3(n + 1)n! - 1}}<br />
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
So the equivalent sum would be
<br /> \sum\limits_{n = 1}^\infty {\left( {\frac{1}{3} + \frac{1}{{3(3n! - 1)}}} \right)} <br />
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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