Where am I going wrong here when applying the Ratio test?

Question

I'm learning the various convergence/divergence test techniques. I'm on Ratio test now. I believe I understand how to apply in that you have some ∑ a_n from n = 1 to ∞ and when applied, it would look like:

lim n->∞ (|a_n+1|/|a_n|

So when solving a problem, first plug in the necessary data, then simplify the expression, cancel terms and you should be left with a limit that can either be greater than, less than, or equal to 1 (Each of these meaning a different outcome). Knowing this, I tried a couple problems. Here is one that I thought I was doing right, but I am a bit confused as to where I am going wrong.

$\sum_{n=1}^∞ \frac{(6n+2)^n}{(5n-4)^{3n}}$

$\frac{\frac{(6(n+1)+2)^{n+1}}{(5(n+1)-4)^{3(n+1)}}}{\frac{(6n+2)^{n}}{(5n-4)^{3n}}}$

$\frac{(6(n+1)+2)^{n+1}}{(5(n+1)-4)^{3(n+1)}}•{\frac{(5n-4)^{3n}}{(6n+2)^{n}}}$

Normally with other examples, I would see obvious terms to cancel, but with this one, I fail to see obvious terms to cancel. Did I do something wrong with setting it up or am I missing something obvious here? Thanks!

Answer 1

We have

$$\frac{(6(n+1)+2)^{n+1}}{(5(n+1)-4)^{3n+1}}{\frac{(5n-4)^{3n}}{(6n+2)^{n}}}=\frac{6(n+1)+2}{5(n+1)-4}\frac{(6(n+1)+2)^{n}}{(6n+2)^{n}}\frac{(5n-4)^{3n}}{(5(n+1)-4)^{3n}}=\\=\frac{6n+8}{5n+1}\left(\frac{6n+8}{6n+2}\right)^n\left(\frac{5n-4}{5n+1}\right)^{3n}=\\=\frac{6n+8}{5n+1}\left(1+\frac{6}{6n+2}\right)^n\left(1-\frac{5}{5n+1}\right)^{3n}$$

and

• $\left(1+\frac{6}{6n+2}\right)^n=\left[\left(1+\frac{6}{6n+2}\right)^{6n+2}\right]^{\frac{n}{6n+2}}\to e$
• $\left(1-\frac{5}{5n+1}\right)^{3n}=\left[\left(1-\frac{5}{5n+1}\right)^{5n+1}\right]^{\frac{3n}{5n+2}}\to\frac1{e^3}$