Divergent Or Convergent Series

[tmt1]

I have this:

$$ \sum_{n = 1}^{\infty} \frac{n^n}{3^{1 + 3n}}$$

And I need to determine if it is convergent or divergent.

I try the limit comparison test against:

$$ \frac{1}{3^{1 + 3n}}$$.

So I need to determine

$$ \lim_{{n}\to{\infty}} \frac{3^{1 + 3n} \cdot n^n}{3^{1 + 3n}}$$

Or

$$ \lim_{{n}\to{\infty}} n^n$$

which is clearly $\infty$.

So that means the initial expression should behave the same as

$$\sum_{n = 1}^{\infty} \frac{1}{3^{1 + 3n}}$$.

Clearly $3^{1 + 3n } > 3^n$, therefore $\frac{1}{3^{1 + 3n }} < \frac{1}{3^n}$

Since $$ \sum_{n = 1}^{\infty} \frac{1}{3^{n}}$$ is convergent, then $$ \sum_{n = 1}^{\infty} \frac{1}{3^{1 + 3n }}$$

is convergent.

Thus, $ \sum_{n = 1}^{\infty} \frac{n^n}{3^{1 + 3n}}$ is convergent.

 

[MarkFL]

I would look at:

\(\displaystyle L=\lim_{n\to\infty}\left(\frac{n^n}{3^{1+3n}}\right)=\frac{1}{3}\lim_{n\to\infty}\left(\left(\frac{n}{27}\right)^n\right)\)

Do we have $L=0$? If not, then by the limit test, the series diverges.