Looking for explanation of bound on Dirichlet's L-Function

Question

I am reading Stein and Shakarchi's Fourier Analysis text and the proof Dirichlet's theorem and I am looking for clarification on how he derives the following for large $s$, $\lim_{s\to\infty}$ and $\chi_q$ is a Dirichlet character for $(\mathbb{Z}/q\mathbb{Z})^*$.

$|L(s,\chi_q) - 1| \le 2q \sum_{n=2}^\infty n^{-s}$

All I can come up with is

$|\chi(n)| \le 1$

then

$|L(s,\chi)| \le \sum_{n=1}^\infty n^{-s}$

and

$|L(s,\chi)| - 1 \le \sum_{n=2}^\infty n^{-s}$

What am I missing here? Where does it come from and is it needed in proving the bound

$L(s,\chi) = 1 + O(2^{-s})$

Answer 1

Ok, I read this section of the book and have an explanation for the $2q$ and it follows along similar lines to what Eric wrote, but I think this argument is more inline with what the book was suggesting for large $s$.

Let $s_k = \sum_{n=2}^{k}\chi(n)$

Then we can rewrite, $\sum_{n=2}^{\infty}\chi(n)n^{-s} = \sum_{n=2}^{\infty}\frac{s_k - s_{k-1}}{n^{s}}$

And since we know that $|s_k| = |\sum_{n=2}^{k}\chi(n)| \le q$

Then $|s_k - s_{k-1}| \le 2q$ and $\sum_{n=2}^{\infty}\frac{s_k - s_{k-1}}{n^{s}} \le \sum_{n=2}^{\infty}\frac{2q}{n^{s}}$

Answer 2

Let $S(t)=\sum_{2\leq n\leq t}\chi(n)$, and suppose that $s>1.$ Using integration by parts, we have that $$\sum_{n=2}^{\infty}\chi(n)n^{-s}=\int_{2}^{\infty}t^{-s}d\left(S(t)\right)=s\int_{2}^{\infty}S(t)t^{-s-1}dt,$$ and by applying the trivial bound $|S(t)|\leq q,$ it follows that $$\left|\sum_{n=2}^{\infty}\chi(n)n^{-s}\right|\leq q2^{-s}.\ \ \ \ \ \ \ \ \ \ (1)$$ Another possible bound, as you noted, is $$\left|\sum_{n=2}^{\infty}\chi(n)n^{-s}\right|\leq\sum_{n=2}^{\infty}n^{-s}\ \ \ \ \ \ \ \ \ \ (2),$$ however, we can ask, does equation $(1)$ have any advantages over equation $(2)$? The answer is yes if we care about $s$ close to $1$. As $s$ approaches $1$, the bound in equation $(2)$ goes to infinity, and in particular, by taking the limit as $s\rightarrow 1$, equation $(1)$ implies that $L(1,\chi)$ is finite.

This is a minor detail, and while I doubt the authors added the $q$ to have a result uniform in $s\in[1,\infty)$, it is likely that they used the above method of partial summation. As you found, for the question of $s\rightarrow \infty$, the trivial bound works best.