If \( f : \mathbb N \to \mathbb C \) is a function, and \( s \in \mathbb C \) with \( \Re(s) > 0 \) is a complex number such that the Dirichlet series \( L(f,s) = \displaystyle\sum_{n=1}^{\infty} \frac{f(n)}{n^s} \) converges,
then \( \displaystyle\sum_{n \le x } f(n) = o(x^s) \quad\text{for}\quad x \to + \infty \)
As an example, take \( f(n) = 1 \), then \( \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{s}} \) converges for \( s > 1 \) and indeed \( \displaystyle\sum_{n \le x } 1 = \lfloor x \rfloor = o(x^s) \) for all \( s > 1 \).
As a second example, take \( f(n) = \frac{1}{n} \), then \( \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1+s}} \) converges for \( s > 0 \) and indeed \( \displaystyle\sum_{n \le x } \frac{1}{n} = \log x = o(x^s) \) for all \( s > 0 \).
What I like about Kronecker's lemma is that its statement and proof involve real analysis only; one does not need to know anything about the behaviour of the function \(L(f,s) \) for complex numbers \( s \). On the other hand, in analytic number theory, one often uses Perron's formula to estimate sums \( \sum_{n \le x } f(n) \) for number theoretic functions \( f \). One then needs to have information about the zeros of the Dirichlet series, and its behaviour for large imaginary values of \( s \) to make Perron's formula work. All this information is thus not needed when using Kronecker's lemma.
Although a proof of Kronecker's lemma can be found in Hildebrand's lecture notes, I write here a proof for the case \( s \in \mathbb R \). This proof is based on the French Wikipedia article about Kronecker's lemma and is very transparent.
