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Let $\chi$ be a non-trivial Dirichlet character modulo $a$. Also assume $a$ has a primitive root $r$.

Prove that Dirichlet $L$ function $L(s,\chi)=\sum_{n=1}^\infty {\chi(n)\over n^{s}}$ converges when $s>0$.

I think the key hint is the primitive root. Please help..

Bhaskar Vashishth
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user234902
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  • This should be discussed in almost any analytic number theory book. Have you read any such texts? – KCd Jun 07 '15 at 07:47
  • @KCd i don't have any analytic number theory book. Can you give me some references about the proof? – user234902 Jun 08 '15 at 04:55
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    Yes, a reference is any analytic number theory book! Just get a few from a library. It doesn't matter if you don't own a book yourself, just use a library to help. And by the way, the condition on $a$ is irrelevant. For every modulus $m \geq 2$, the Dirichlet $L$-function of every nontrivial character $\chi \bmod m$ converges for $s > 0$. – KCd Jun 08 '15 at 05:47
  • Sorry, but are you attending an "introduction to the number theory" course in some school in KAIST? – dust05 Jun 09 '15 at 05:37

1 Answers1

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There are two key steps: bounding the partial sums of the Dirichlet character, and determining the region of analyticity of its L-function.

First, consider the sum $$S(a)=\sum_{n=1}^a \chi(n).$$ Since $\chi$ is not the trivial or principal character, there exists an integer $b$ between $1$ and $a$ such that $b$ is coprime to $a$ and $\chi(b)\neq 1$. Therefore $$\chi(b)S(a)=\sum_{n=1}^a \chi(b)\chi(n)=\sum_{n=1}^a \chi(bn)$$ by the complete multiplicity of $\chi$. Since the summation is such that there is exactly one $n$ congruent to each possible residue modulo $a$, there is also exactly one $bn$ congruent to each possible residue modulo $a$ for a given integer $b$; therefore this second sum is precisely $S(a)$. We then have $$\chi(b)S(a)=S(a),$$ and since $\chi(b)\neq 1$, we have $S(a)=0$. Since $\chi$ is periodic, it follows that $$S(N)=\sum_{n=1}^N \chi(n)=\sum_{n=a\lfloor N/a\rfloor}^N \chi(n)<a=O(1). \tag 1$$

Now consider the L-function for this character: $$L(s,\chi)=\sum_{n=1}^\infty \frac{\chi(n)}{n^s}=\int_1^\infty \frac{S(x)}{x^{s+1}}\; dx \tag 2$$ by Abel's summation formula (partial summation). We note that $S(x)$ exists for all $x \in [1,\infty)$. In order for the integral in $(2)$ to diverge, we would need, for instance, $\frac{S(x)}{x^{s+1}}\gg \frac 1{x \log^2 x}$ and thus $S(x)\gg \frac{x^s}{\log^2 x}$. However, by $(1)$ this is false for any $s$ with $\Re(s)>0$. Therefore $L(s,\chi)$ converges for any $\Re(s)>0$.

For further reference on classical analytic number theory, I recommend these notes by A. J. Hildebrand.

Avi
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