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..
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.