If $n\in\mathbb N$, then $\sum_{k=1}^{\infty} \frac{1}{k^n}$ diverges. What's wrong in my proof?

Question

If $n\in\mathbb N$, then $\sum_{k=1}^{\infty} \frac{1}{k^n}$ diverges.

Clearly this is false in general and only holds for $n=1$ . But here's my wrong proof in which I can't find what's wrong.

Let $$S_n=\sum_{k=1}^{\infty}\frac{1}{k^n}$$.

Now , $k$ is a natural number, hence it can be represented as a product of primes raised to some whole number power as per fundamental theorem of arithmetic.

Hence,

$$S_n= \cdots\sum_{a_3≥0}\sum_{a_2≥0}\sum_{a_1≥0}\frac{1}{(2^{a_1}3^{a_2}5^{a_3}...)^n}$$

Hence,

$$S_n=\sum_{a_1≥0}\frac{1}{2^{na_1}}\sum_{a_2≥0}\frac{1}{3^{na_2}} \sum_{a_3≥0}\frac{1}{5^{na_3}}\cdots$$

Now , each sum in multiplication is an infinite geometric progression. It's easy to see that each sum is greater than $1$ as follows :

$$\sum_{a_i≥0}\frac{1}{(p_i)^{na_i}}= \frac{{p_i}^n}{{p_i}^n-1}≥1$$

where $p_i$ is the $i^{th}$ prime number.

Hence , $S_p$ is a product of infinite numbers where each of these numbers is greater than 1 , hence it diverges.

Thanks , please tell where did I go wrong in this proof.

Answer 1

Aside from the significant issue pointed out by Sangchul in the comments, there is also a second more technical issue: you are trying to show that the infinite sum diverges, meaning you are trying to show that the sequence of partial sums

$$\sum_{k=1}^N \frac{1}{k^n}$$

does not have a limit as $N \to \infty$. But when you rewrite the infinite sum as an infinite product (which doesn't obviously make sense unless it converges) and try to argue that the infinite product diverges, you are arguing (incorrectly) that the sequence of partial products

$$\prod_{i=1}^M \left( \frac{1}{1 - p_i^{-n}} \right)$$

does not have a limit as $M \to \infty$. It actually does, but even if it didn't, that doesn't directly imply that the above sequence of partial sums diverges, because the two aren't the same sequence!

Instead a rigorous form of this strategy would have been to proceed by showing that $\prod_{i=1}^M \left( \frac{1}{1 - p_i^{-n}} \right)$ is a lower bound on the infinite sum, because it corresponds to the sum over all $k$ divisible by only the first $M$ primes, then show that it diverges to $\infty$ as $M \to \infty$ (which, again, it doesn't when $n > 1$). This infinite product actually does diverge when $n = 1$ but this is actually harder to show than the divergence of the harmonic series; via taking the logarithm it's equivalent to the divergence of the series $\sum_i \frac{1}{p_i}$. In fact the standard approach is to reverse this argument; it's possible to show that the divergence of the harmonic series implies the divergence of this series.