$\displaystyle a_n = \frac{ (6 n + 6) 6^{n + 3} }{ 4^{n} }
$
Can someone please explain the steps of solving this
-Thanks
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.
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: $\displaystyle = \lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot 4^{n-(n+1)} \cdot 6^{n+4 - (n+3)}\right| = \lim_{n \to \infty} \left| \frac{6n+12}{6n+6} \cdot \frac{6}{4}\right| = \cdots$
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?
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.