I'm trying to prove that the Dirichlet density of the set $X$ of primes of the form $p = n^2+1$ is zero. The only way I can think to do it is to put a bound on the sum $\Sigma_{p\in X} \frac{1}{p^s}$, that way when I take the limit as $s\rightarrow 1^+$ the sum will converge and the density will be zero.
The furthest I've gotten is showing that the truncated sum is less than the partial sum $\Sigma \frac{1}{m^s}$. This doesn't help much though because as we take the upper limit of each sum to infinity we get (assuming $s>1$) $\Sigma_{p\in X} \frac{1}{p^s}<\zeta (s)$, but the zeta function diverges as $s\rightarrow 1^+$. Besides, you can do this for any set of primes I think .
Edit: Is this approach valid?: Let $N=\{n_i\}_{i\in I}$ be the set of integers $n_i$ making $n_i^2+1$ prime. $\Sigma _{i=1}^M \frac{1}{(n_i^2 + 1)^s}<\Sigma_{j=1}^{M}\frac{1}{j^{2s}}$. Taking the limit as $M\rightarrow \infty$ (which is OK for $s>1/2$), we see that $\Sigma_{p\in X} \frac{1}{p^s}$ is bounded for $s>1/2$, which means the limit as $s\rightarrow 1^+$ is finite?
Thank you
Your proof is correct. Rewriting your argument in a little clear way (assume $s$ real, $s>1/2$ )$$\sum_{\underset{{\scriptstyle p\in X}}{p\leq M}}\frac{1}{p^{s}}\leq\sum_{n\leq M}\frac{1}{\left(n^{2}+1\right)^{s}}\leq\sum_{n\leq M}\frac{1}{n^{2s}}$$ so if $M\rightarrow\infty$ $$\sum_{{\scriptstyle p\in X}}\frac{1}{p^{s}}\leq\zeta\left(2s\right)$$ and if $s\rightarrow1$ $$\sum_{{\scriptstyle p\in X}}\frac{1}{p}\leq\zeta\left(2\right)=\frac{\pi^{2}}{6}$$ so well done. Now is clear that $$\lim_{s\rightarrow1^{+}}\frac{\sum_{{\scriptstyle p\in X}}\frac{1}{p^{s}}}{\log\left(\frac{1}{s-1}\right)}=0.$$