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.
On
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.
On
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.
On
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).
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 $\infty$.