$d (A) = \lim_{\sigma \to 1^+}\frac{1}{\zeta(\sigma)}\sum_{n \in B} \frac{1}{n^{\sigma}}$ for $B \subset \Bbb{N}$. So clearly this limit is $0$ for reciprocally summable (convergent) $B$.
My goal is to prove that $\frac{1}{\zeta(\sigma)}$ can be replaced with $\sigma - 1$ in that formula using Abel Partial Summation.
My understanding is that the limit is taken "from the right" over the reals because of the $\to 1^+$.
Abel's Partial Summation is:
For any divergent sequence $\lambda_n$ of real numbers. Let $f(x)$ be a $\Bbb{C}$-valued continuous, differentiable function & let $A(x) = \sum_{\lambda_n \leq x} \alpha_n$ for some complex numbers $\alpha_n$. Then: $$\sum_{\lambda_n \leq x} \alpha_n f(\lambda_n) = A(x)f(x) - \int_{\lambda_1}^x A(u)f'(u) \, {\rm d}u$$
What I've got so far is: If $B$ is finite, then this is clearly true as then the limit with $\sigma - 1$ substituted in is $0$.
So let $B$ be infinite, then the divergent sequence we need is staring us in the face. Let $\lambda_1 \lt \lambda_2 \lt \dots = B$.
Fix $\sigma$ and let $f(x) = \dfrac{1}{x^{\sigma}}$. Define $A(x) = \sum_{\lambda_n \leq x} 1$ ie. $\alpha_n = 1$ for all $n$. Then by Abel's partial summation we have:
$$ \sum_{\lambda_n \leq x} f(\lambda_n) = \sum_{\lambda_n \leq x} \frac{1}{n^\sigma} = \frac{\sum_{\lambda_n \leq x} 1}{x^{\sigma}} + \int_{1}^x \frac{\sigma \sum_{\lambda_n \leq u} 1}{u^{\sigma + 1}} \, {\rm d}u.$$
Then I'm stuck.