Testing for convergence of expression with variable denominator raised to some constant 'a'

Question

I am practicing some convergence test problems for an exam and encountered a problem I am not sure how to approach: $$\sum_{n=1}^{\infty}\frac{1}{n^{\ln3}}$$

My approach was the following:

I used the ratio test: $$\lim_{n\rightarrow\infty}{\frac{1}{({n+1})^{\ln3}}\div{\frac{1}{n^{\ln3}}}}$$

Since $\ln3$ is just some arbitrary constant $a>1$, I assumed I could just choose some other constant that would fit the description. I used $a=2$.

$$\Rightarrow \lim_{n\rightarrow\infty}{\frac{1}{({n+1})^{2}}\div{\frac{1}{n^{2}}}}$$

Hence,

$$\lim_{n\rightarrow\infty}{\frac{1}{({n+1})^{2}}\div{\frac{1}{n^{2}}}}=\lim_{n\rightarrow\infty}{\frac{n^{2}}{({n+1})^{2}}}=1$$

I could go a step further and prove it works for any $a\in\Bbb{Z}>0$ because it's just a binomial expansion where the numerator is the leading term of the denominator (i.e. it's $\lim_{n\rightarrow\infty}=1$). Now, I am not sure about the cases where $a\notin\Bbb{Z}$ but my assumption is that the leading term of the denominator is always also the numerator and thus $\lim_{n\rightarrow\infty}=1$.

My question is: is my approach valid? And is there a better way to do this?

Edit: Ok, I realized that even if my methodology were correct, that leads me nowhere because if the ratio test gives $1$, it doesn't prove convergence or divergence. But, my question still stands as to whether I did the procedure right or wrong.

Answer 1

I could answer it using the integral test:

$\sum_{n=1}^\infty\frac{1}{n^{\ln3}}$

$\Rightarrow\frac{1}{(n+1)^{\ln3}}\le\frac{1}{n^{\ln3}},n\ge1$

Then,

$\int_1^\infty\frac{1}{n^{\ln3}}=\lim_{n\rightarrow\infty}\frac{1}{(1-\ln3)n^{\ln3-1}}=0$

Therefore, $S_n$ converges.

Likewise, you can prove the $p$-series for $p\le1$.

Answer 2

Your series represents what we call genearlized harmonic serie that is defined as follows:

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

Using Mengoli's serie, we can show that: $\alpha\in(1,+\infty)$ converges.

$\alpha\in (-\infty, 1]$ diverges.

Now, you can use the comparison test, to show that you series converges.

Comparison test. Let $a_n$ and $b_n$ two real sequences definitely non-negative and such that $1\leq a_n\leq b_n$. The following expression are true:

If $\sum a_n$ diverges, than also $\sum b_n$ diverges.

If $\sum b_n$ converges, then also $\sum a_n$ converges.