Ratio Test

**fastcarslaugh** · November 17th 2008, 06:50 PM

$a_n = \frac{ (6 n + 6) 6^{n + 3} }{ 4^{n} }<br />$

Can someone please explain the steps of solving this

-Thanks

**o_O** · November 17th 2008, 07:02 PM

If $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$ , then $\sum a_n$ is absolutely convergent.

So: $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left| {\color{red}a_{n+1}} \cdot {\color{blue}\frac{1}{a_n}} \right| = \lim_{n \to \infty} \left| {\color{red}\frac{\left(6(n+1) + 6\right)6^{(n+1)+3}}{4^{n+1}}} \cdot {\color{blue}\frac{4^n}{(6n + 6)6^{n+3}}}\right| = \hdots$

See where you can go from here. Post back any specific problems you may have.

**fastcarslaugh** · November 17th 2008, 07:13 PM

I guess I asked a wrong question there

I already got to that part it's just that I am not sure about canceling the numbers

**o_O** · November 17th 2008, 07:21 PM

Well simplifying a bit we get: $\lim_{n \to \infty} \left| \frac{\left(6n+ 12\right)6^{n+4}}{4^{n+1}} \cdot \frac{4^n}{(6n + 6)6^{n+3}}\right|$

Make use of the property that: $\frac{a^m}{a^n} = a^{m-n}$

So we get: $= \lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot 4^{n-(n+1)} \cdot 6^{n+4 - (n+3)}\right| = \lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot \frac{6}{4}\right| = \cdots$

**fastcarslaugh** · November 17th 2008, 07:34 PM

So I realized that I was able to simplify this correctly.

so I pull out $3/2$ then $\lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot \right|$

so top is approaching infinity faster than the bottom. Does this mean that it's 1?

So the limit is $3/2$ and the ratio test is inconclusive. But will the series converge or diverge by another test?

**Mathstud28** · November 17th 2008, 07:47 PM

Originally Posted by fastcarslaugh

So I realized that I was able to simplify this correctly.

so I pull out $3/2$ then $\lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot \right|$

so top is approaching infinity faster than the bottom. Does this mean that it's 1?

So the limit is $3/2$ and the ratio test is inconclusive. But will the series converge or diverge by another test?

It is not inconclusive, if the ratio test yields a number greater than one it is divergent. If you'd care for further proof of its divergence why not try the n-th term test for divergence?

**fastcarslaugh** · November 17th 2008, 07:56 PM

For some random reason, the program that I use to do my homework would not accept the correct answer. After 37th times, it finally accepted the answer. and I believe that I typed the same correct answer about 34 times.

anyway thanks for clearing my doubt.

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