9
$\begingroup$

The sum $\sum_{n=1}^{\infty} 1/n^{s}$ is convergent for all real $s>1$ and diverges for all real $s \le 1$. The same holds for the sum $\sum_{p \ prime} 1/p^{s}$. Thus, for the functions $f(n)= 1/n^s, s \in \mathbb{R}$ the sum $\sum_{n=1}^{\infty}f(n)$ shows the same convergence behaviour as the sum $\sum_{p \ prime}f(p)$.

The same holds, if I'm not mistaken, for the functions $f(n)= 1/(n (\ln n)^s), s \in \mathbb{R}$ (both for $n \in \mathbb{N}$ and for primes convergence iff $s>1$).

Question: Is there a real monotonic function $f$ such that $\sum_{n=1}^{\infty}f(n)$ diverges whereas sum $\sum_{p \ prime}f(p)$ converges? (The monotony requirement is for preventing 'artificial' solutions that single out the primes (as e.g. $f(n) = 2^n$ if $n$ is prime; $f(n)=n$ if $n$ is not prime)).

$\endgroup$
5
  • 4
    $\begingroup$ The simple approach to this problem is to replace \sum_p f(p) with the roughly equivalent \sum_n f(n log n) since the nth prime is roughly n log n. Once you've found a monotonic function where \sum_n f(n log n) converges but \sum_n f(n) diverges then it probably won't be too hard to use the prime number theorem to answer your question. $\endgroup$ Jul 6, 2010 at 21:12
  • 3
    $\begingroup$ Or, easier to think about, a function $f(t)$ such that $\int f(t) dt$ converges and $\int f(t) dt/\log t$ diverges. $\endgroup$ Jul 6, 2010 at 21:15
  • 4
    $\begingroup$ f(n) = n log n has that property, doesn't it? $\endgroup$ Jul 6, 2010 at 21:22
  • $\begingroup$ Switch converges and diverges in my comment. $\endgroup$ Jul 6, 2010 at 21:24
  • 1
    $\begingroup$ Qiaochu, I think you mean $f(n) = 1/(n \log n)$. $\endgroup$ Jul 6, 2010 at 22:05

2 Answers 2

11
$\begingroup$

I think, you are mistaken, sum $\sum 1/(p\log p)$ converges, since $p_n\log p_n$ behaves like $n(\log n)^2$

$\endgroup$
1
  • $\begingroup$ Thank you, how could I have overlooked this? $\endgroup$ Jul 6, 2010 at 21:44
5
$\begingroup$

Note also that if $\{a_n\}$ is an increasing sequence of natural numbers that presents arbitrarily large gaps between consecutive terms (e.g. a sequence with with density $0$), there is always a positive decreasing function $f$ such that the sum of $f(a_n)$ converges and the sum of $f(n)$ diverges to positive infinity. The reason is that $\sum_n f(n)\geq\sum_n f(a_n)(a_n-a_{n-1}),$ and as the $a_n-a_{n-1}$ are unbounded, the claim reduces to the easy: for any unbounded sequence $u_n>0$ there is a convergent series with coefficients $\epsilon_n>0$ such that the series of the $u_n\epsilon_n$ diverges. Use this taking $u_n:=a_n-a_{n-1}$ to define a decreasing $f$ such that $f(a_n)=\epsilon_n$.

(Just to clarify that if your question was really "is there a function" and not "which function", then it has a basic answer, not requiring the prime number theorem).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.