Verify that ## L(1, \chi)\neq 0 ##

[Math100]

Homework Statement

For each real-valued nonprincipal character $\chi$ mod $16$, verify that $L(1, \chi)\neq 0$.

Relevant Equations

$L(1, \chi)=\sum_{n=1}^{\infty}\frac{\chi(n)}{n}$
Note that $L(1, \chi)\neq 0$ when $\chi$ is a nonprincipal character.

From a table of values for all the nonprincipal Dirichlet characters modulo $16$, we have that
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline \chi_{2}(n) & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 \\
\hline \chi_{3}(n) & 1 & i & i & 1 & -1 & -i & -i & -1 \\
\hline \chi_{4}(n) & 1 & -i & i & -1 & -1 & i & -i & 1 \\
\hline \chi_{5}(n) & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 \\
\hline \chi_{6}(n) & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 \\
\hline \chi_{7}(n) & 1 & i & -i & -1 & -1 & -i & i & 1 \\
\hline \chi_{8}(n) & 1 & -i & -i & 1 & -1 & i & i & -1 \\
\hline
\end{array}

So $L(1, \chi)=\sum_{n=1}^{\infty}\frac{\chi_{k}(n)}{n}$ with $k\in\left \{ 2, 5, 6 \right \}$.
Observe that
\begin{align*}
&\sum_{n=1}^{\infty}\frac{\chi_{2}(n)}{n}=\frac{\chi_{2}(1)}{1}+\frac{\chi_{2}(3)}{3}+\frac{\chi_{2}(5)}{5}+\frac{\chi_{2}(7)}{7}+\frac{\chi_{2}(9)}{9}+\frac{\chi_{2}(11)}{11}+\frac{\chi_{2}(13)}{13}+\frac{\chi_{2}(15)}{15}\\
&1+(-\frac{1}{3})+\frac{1}{5}+(-\frac{1}{7})+\frac{1}{9}+(-\frac{1}{11})+\frac{1}{13}+(-\frac{1}{15})=\frac{33976}{45045}\neq 0\\
&\sum_{n=1}^{\infty}\frac{\chi_{5}(n)}{n}=\frac{\chi_{5}(1)}{1}+\frac{\chi_{5}(3)}{3}+\frac{\chi_{5}(5)}{5}+\frac{\chi_{5}(7)}{7}+\frac{\chi_{5}(9)}{9}+\frac{\chi_{5}(11)}{11}+\frac{\chi_{5}(13)}{13}+\frac{\chi_{5}(15)}{15}\\
&1+\frac{1}{3}+(-\frac{1}{5})+(-\frac{1}{7})+\frac{1}{9}+\frac{1}{11}+(-\frac{1}{13})+(-\frac{1}{15})=\frac{47248}{45045}\neq 0\\
&\sum_{n=1}^{\infty}\frac{\chi_{6}(n)}{n}=\frac{\chi_{6}(1)}{1}+\frac{\chi_{6}(3)}{3}+\frac{\chi_{6}(5)}{5}+\frac{\chi_{6}(7)}{7}+\frac{\chi_{6}(9)}{9}+\frac{\chi_{6}(11)}{11}+\frac{\chi_{6}(13)}{13}+\frac{\chi_{6}(15)}{15}\\
&1+(-\frac{1}{3})+(-\frac{1}{5})+\frac{1}{7}+\frac{1}{9}+(-\frac{1}{11})+(-\frac{1}{13})+\frac{1}{15}=\frac{27904}{45045}\neq 0.\\
\end{align*}
Thus
\begin{align*}
&L(1, \chi)=\sum_{n=1}^{\infty}\frac{\chi_{2}(n)}{n}+\sum_{n=1}^{\infty}\frac{\chi_{5}(n)}{n}+\sum_{n=1}^{\infty}\frac{\chi_{6}(n)}{n}\\
&=\frac{33976}{45045}+\frac{47248}{45045}+\frac{27904}{45045}\\
&=\frac{36376}{15015}\\
&\neq 0.\\
\end{align*}
Therefore, $L(1, \chi)\neq 0$ for each real-valued nonprincipal character $\chi$ mod $16$.

 

[fresh_42]

You have $\chi\in \{\chi_2,\chi_5,\chi_6\}.$ $L(1,\chi)$ is not summed over all non-principal characters so everything from "Thus" on must be deleted. You have to calculate $L(1,\chi_2),L(1,\chi_5),L(1,\chi_6)$ which you did.

The rest is ok. Maybe it's worth mention that $L(1,\chi)\in \mathbb{C}.$

 

[Math100]

You have $\chi\in \{\chi_2,\chi_5,\chi_6\}.$ $L(1,\chi)$ is not summed over all non-principal characters so everything from "Thus" on must be deleted. You have to calculate $L(1,\chi_2),L(1,\chi_5),L(1,\chi_6)$ which you did.

The rest is ok. Maybe it's worth mention that $L(1,\chi)\in \mathbb{C}.$

So it's asking for the summation for each of the $\chi_{2}, \chi_{5}, \chi_{6}$ then?

 

[fresh_42]

So it's asking for the summation for each of the $\chi_{2}, \chi_{5}, \chi_{6}$ then?

No. One sum per character, makes three sums. Everything was fine until you started with "Thus".

\begin{align*}
L(1,\chi_2)&=\sum_{n=1}^\infty \dfrac{\chi_2(n)}{n}=\ldots \neq 0\\
L(1,\chi_5)&=\sum_{n=1}^\infty \dfrac{\chi_5(n)}{n}=\ldots \neq 0\\
L(1,\chi_6)&=\sum_{n=1}^\infty \dfrac{\chi_6(n)}{n}=\ldots \neq 0
\end{align*}

 

[Math100]

No. One sum per character, makes three sums. Everything was fine until you started with "Thus".

\begin{align*}
L(1,\chi_2)&=\sum_{n=1}^\infty \dfrac{\chi_2(n)}{n}=\ldots \neq 0\\
L(1,\chi_5)&=\sum_{n=1}^\infty \dfrac{\chi_5(n)}{n}=\ldots \neq 0\\
L(1,\chi_6)&=\sum_{n=1}^\infty \dfrac{\chi_6(n)}{n}=\ldots \neq 0
\end{align*}

I see now.