No squares allowed

Calculus Level 2

Any integer $n$ can be uniquely written as a product of the form $n = s^2 m$ . For example, $18 = 3^2 \cdot 2$ . Square-free integers are integers such that $s=1$ , so they are not divisible by any perfect square.

For the set of all square-free integers $S$ , evaluate

$\displaystyle\sum_{m \in S}^\infty \frac{1}{m}$

If the sum diverges, input $-1$ .

The answer is -1.

2 solutions

Levi Walker
Nov 25, 2018

Notice that every prime number, $p$ , is squarefree. Thus, $\displaystyle \sum_{m=1}^\infty \frac{1}{m} > \displaystyle \sum_{p=1}^\infty \frac{1}{p}$

The right hand side diverges (see this post , for the proof). We then have

$\infty < \displaystyle \sum_{m=1}^\infty \frac{1}{m}$

Thus the sum in question diverges as well. So the answer is $\boxed{-1}$ .

You don't need an upper "bound" here (also, infinity is not a bound!) - the second half of your proof (together with the observation that every term in the sum is positive) is enough.

Chris Lewis - 2 years, 6 months ago

Otto Bretscher
Nov 25, 2018

Assuming that $\sum_{m \in S}\frac{1}{m}$ converges (where $S$ denotes the square-free numbers), we must conclude that $(\sum_n\frac{1}{n^2}) (\sum_{m \in S}\frac{1}{m})=\sum_n\frac{1}{n}$ converges too, a contradiction.

Thus the answer is $\boxed{-1}$

No squares allowed

The answer is -1.

2 solutions

1 pending report

Vote up reports you agree with