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Given that $$ \sum_{n \ge 1} \frac{a_n}{n^s} = \sum_{n \ge 1} \frac{b_n}{n^s}, $$ for all complex values of $s$ and where $a_n$ and $b_n$ are non-negative integers, can we say that $a_n = b_n$ for all $n$?

Genesis of the question: While studying the basics of Gaussian integers, I came across such an equation in https://www.maths.nottingham.ac.uk/plp/pmzcw/download/fnt_chap5.pdf (page 7).

Ideas: Of course if you replace the denominators $\{1^s, 2^s, 3^s, \dots\}$ with variables $\{x_1, x_2, x_3,\dots\}$ or $\{x, x^2, x^3, \dots\}$ then, it will be true. But we cannot do that.

I don't think induction will work.

Feng
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Arpita Korwar
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1 Answers1

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Yes, it's true. $a_n$ and $b_n$ can be complex numbers and in fact you only need the identity to hold for $s$ in some open subset of $\mathbb{C}$, and maybe even less than that. But since you allow all $s$ (I will only need sufficiently large real $s$) we can simplify the argument as follows.

Take $s \to \infty$ to be real. The limit on each side is $a_1$ and $b_1$ respectively, so we get $a_1 = b_1$. Each term $n^{-s}$ vanishes exponentially faster than the previous term so the two sides grow as $a_1 + a_2 2^{-s} + O(3^{-s})$ resp. $b_1 + b_2 2^{-s} + O(3^{-s})$ and matching growth rates gives $a_2 = b_2$. Then we can continue matching up growth rates on each side term by term. (I am being sloppy here; we need to bound the $a_n$ and the $b_n$ using the fact that the Dirichlet series converges to do this more carefully.)

As an alternative argument you can use Perron's formula.

Qiaochu Yuan
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  • Then we can ask more. Suppose the equation is true on and interval in $\mathbb R$. Or true on a set with a limit point. – GEdgar Jul 19 '22 at 11:54
  • @GEdgar: right, so I think we can bootstrap as follows: if the two Dirichlet series converge anywhere then they both converge and are holomorphic on some half-plane ${ \text{Re}(s) > r }$. Then if they agree on a subset of this half-plane with a limit point we can apply the identity theorem to conclude that they agree on the entire half-plane, then run the above argument. – Qiaochu Yuan Jul 19 '22 at 12:09