Determine if the SERIES converges or DIVERGES(II)

[shamieh]

$$\sum^{\infty}_{n = 0} \frac{3^{2n + 1}}{n5^{n-1}}$$ using Ratio test I obtained: $$\frac{3^{2n + 2}}{(n + 1)5^n} * \frac{n5^{n - 1}}{3^{2n + 1}}$$ = 25/2 so the series diverges since L > 1?

 

[Prove It]

$$\sum^{\infty}_{n = 0} \frac{3^{2n + 1}}{n5^{n-1}}$$ using Ratio test I obtained: $$\frac{3^{2n + 2}}{(n + 1)5^n} * \frac{n5^{n - 1}}{3^{2n + 1}}$$ = 25/2 so the series diverges since L > 1?

First of all $\displaystyle \begin{align*} a_n = \frac{3^{2n + 1}}{n \cdot 5^{n - 1}} \end{align*}$, so $\displaystyle \begin{align*} a_{n + 1} = \frac{3^{2 \left( n + 1 \right) + 1 }}{ \left( n + 1 \right) \cdot 5^{n + 1 - 1 } } = \frac{3^{2n + 3}}{\left( n + 1 \right) \cdot 5^n} \end{align*}$

so the ratio is actually

$\displaystyle \begin{align*} \frac{a_{n + 1}}{a_n} = \frac{3^{2n + 3}}{ \left( n + 1 \right) \cdot 5^n } \cdot \frac{n \cdot 5^{n - 1}}{3^{2n + 1}} \end{align*}$

Now see what happens to this ratio when $\displaystyle \begin{align*} n \to \infty \end{align*}$.

 

[shamieh]

so as n--> $$\infty$$ I got 15...? Would that be correct? If not, then i have NO idea how to do algebra and I will need you to show me step by step.

 

[Prove It]

so as n--> $$\infty$$ I got 15...? Would that be correct? If not, then i have NO idea how to do algebra and I will need you to show me step by step.

15 isn't even close (I get 9/5). Please show me what you did.

 

[shamieh]

$$
\frac{3^{2n + 2}}{2^{2n + 1}} = \frac{(3^2)(3^n)(3^2)}{(3^2)(3^n)(3^1)} = 9/3$$

then

for the other part

$$\frac{n * (5^n) (5^1)}{n + 1 (5^n)} = 5/1$$

 

[Prove It]

$$
\frac{3^{2n + 2}}{2^{2n + 1}} = \frac{(3^2)(3^n)(3^2)}{(3^2)(3^n)(3^1)} = 9/3$$

then

for the other part

$$\frac{n * (5^n) (5^1)}{n + 1 (5^n)} = 5/1$$

Yes, the first part is 9/3 = 3.

As for the second part, you can't just treat the n's like they don't exist, they have a part to play. But I'll assume you realized that their limit is 1 and so are negligible.

Also, where have the extra 5's come from? $\displaystyle \begin{align*} \frac{5^{n - 1}}{5^n} = 5^{n - 1 - n} = 5^{-1} = \frac{1}{5} \end{align*}$...

 

[zzephod]

$$\sum^{\infty}_{n = 0} \frac{3^{2n + 1}}{n5^{n-1}}$$ using Ratio test I obtained: $$\frac{3^{2n + 2}}{(n + 1)5^n} * \frac{n5^{n - 1}}{3^{2n + 1}}$$ = 25/2 so the series diverges since L > 1?

You will find this easier to do if you rearrange the $$n$$ th term to be:

$$a_n= \frac{3^{2n + 1}}{n5^{n-1}}=15\left(\frac{9}{5}\right)^n\frac{1}{n}$$

when you will see that you have no term when $$n=0$$, and that $$\lim_{n \to \infty}a_n \ne 0$$