Question: We know that $$ \sum\_{n=1}^{\infty} \frac {1}{n}$$  

diverges to $\infty$. We also know that 
$$\sum\_{n=1}^{\infty} \frac {1}{n^2}$$ 

converges to $\frac {\pi^2}{6}$. 

Is there any $\alpha$ so powerful that the series $$\sum\_{n=1}^{\infty} \frac {1}{\alpha \cdot n}$$ 

converges ? Is there any $\alpha$ such that $\alpha \cdot n$ is nearly as powerful as doing $n^2$ ?  

What is the lowest $\alpha$ such that the series converges ?  

Thanks.

Answer: 
For any $\epsilon > 0$, we have that $\sum\frac{1}{n^{1+\epsilon}}$ converges. In other words: if, as $n$ increases, $\alpha$ is comparable to any root of $n$, for instance if $\alpha = \sqrt[1000]n$, then the series converges.

However, it is also known that by making $\epsilon$ small enough, we can make the sum as large as we want. Therefore, if $\alpha$ is (eventually) smaller than any root (for instance, if $\alpha$ is a constant or $\ln n$, or $a(\ln n)^b$ with $a, b$ constants, for that matter), then $\sum \frac{1}{\alpha n}$ diverges to $\infty$.
The answer to your question is no, because the integral test shows it:

$$\int\_1^{\infty}\frac {1}{\alpha x}dx=\frac {1}{\alpha}\int\_1^{\infty}\frac {1}{x}dx=+\infty$$

You could ask if there is any $1<\alpha(n)<n$ such that $\sum\_{n=1}^{\infty} \frac {1}{\alpha(n) \cdot n}$ converges?

I think the bordering function has not been found yet!

Note: The series $\sum\_{n=1}^{\infty} \frac {1}{n\cdot \ln n}$, $\sum\_{n=1}^{\infty} \frac {1}{n\cdot \ln n \cdot \ln(\ln n)},etc$ also diverge (you can check by the integral test).