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 $\infty$.
If you mean any constant finite $\alpha$, then the short answer is no. You can just take it outside of the summation, and will get the product of $\frac{1}{\alpha}$ and a diverging series.