# uniform convergence of $L(s,\chi)$ for $\Re(s) ≥ 1 + \delta$ " due to absolute convergence for $\Re(s)>1$?

On page 6 of this link, lemma 2.4 shows $L(s,\chi)$ is absolutely convergent for $\Re(s)>1$.

I understand the proof. However, they also add: "The above proof also shows that for any $\delta > 0$, the above series converges absolutely and uniformly for $\Re(s) ≥ 1 + \delta$ "

I do not see why this is the case.

• Note $|n^{-s}|\leq n^{-(1+\delta)}$ for $\Re s\geq 1+\delta$. – Wojowu Oct 15 '17 at 18:21
• @Wojowu and why does this imply uniform convergence? – usere5225321 Oct 15 '17 at 18:40
• Weierstrass M-test. – Wojowu Oct 15 '17 at 18:42
• Cool. So here I would let $f_n(x)= \chi(n)/n$ and I can write $|f_n(x)| \leq |n^{-s}| \leq n^{-(1+\delta)}$ for $\Re(s)>1+\delta$ – usere5225321 Oct 15 '17 at 18:52
• @Wojowu, and we know $\sum n^{-(1+\delta)}$ converges – usere5225321 Oct 15 '17 at 18:53