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: 
By definition:

$$\begin{align}\sum\_{n=1}^\infty\frac1{\alpha n}&\equiv\lim\_{N\to\infty}\sum\_{n=1}^N\frac1{\alpha n}\\&=\frac1\alpha\lim\_{n\to\infty}\sum\_{n=1}^N\frac1n\\&=\infty\cdot\operatorname{sgn}(\alpha)\end{align}$$

So the only impact $\alpha$ has is on the signs.
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).