# Thread: Ratio Test

1. ## Ratio Test

$\displaystyle a_n = \frac{ (6 n + 6) 6^{n + 3} }{ 4^{n} }$

Can someone please explain the steps of solving this

-Thanks

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

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

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

3. 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

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

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

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

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

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

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

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

6. Originally Posted by fastcarslaugh
So I realized that I was able to simplify this correctly.

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

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

So the limit is $\displaystyle 3/2$ 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?

7. 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.