Want to show $\sum_{n>N}n^{-1/2} \ll N^{-1/2}$ I feel like I can use the result $$\sum_{n=x}^N \frac{a_n}{n^s} = A(N)N^{-s} + s \int_x^N A(t)t^{-s-1}dt$$ where $s=1/2$ to verify the $\ll$ approximation.
If I pick $A(n)=\sum_{x\leq t\leq n}\chi(t)$, then I know $|A(n)|\ll 1$. I think that will make the first term become $O(N^{-1/2})$, but am not sure how I should go about handling the integral term. Thanks!
Thanks for the help in advance!
 A: $$\sum_{n=x}^N \frac{a_n}{n^s} = A(N)N^{-s} + s \int_x^N A(t)t^{-s-1}dt$$
gives for $\Re(s) >0$ $$|\sum_{n=x}^\infty \frac{\chi(n)}{n^s}| =| s \int_x^\infty (\sum_{x\leq n\leq t}\chi(n))t^{-s-1}dt|\le  |s| \int_x^\infty q t^{-\Re(s)-1}dt= \frac{q|s| x^{-\Re(s)}}{\Re(s)}$$
A: You have to edit the title as I think you actually meant $\sum_{n>N} \chi(n) n^{-1/2} \ll N^{-1/2} $. While this is valid for a Dirichlet character that is non-principal, it fails when $\chi$ is principal. Letting $$ S(y):= \sum_{1\leq n \leq y } \chi(n) $$ we get $$ \sum_{1\leq n \leq x}\chi(n) n^{-1/2} =\frac{S(x)}{x^{1/2}}+ \frac{1}{2} \int_1^x \frac{S( u ) }{u^{3/2 } } \mathrm d u.$$ Using this for $x=N$ and $x=M$ and subtracting we get $$ \sum_{N< n \leq M}\chi(n) n^{-1/2} =\frac{S(M)}{M^{1/2}}-\frac{S(N)}{N^{1/2}}+ \frac{1}{2} \int_N^M \frac{S( u ) }{u^{3/2 } } \mathrm d u.$$ By the estimate $S(u ) =O(1)$ this  becomes $$\sum_{N< n \leq M}\chi(n) n^{-1/2} = O\left( \frac{1 }{M^{1/2}} \right)
-\frac{S(N)}{N^{1/2}}
+  \frac{1}{2} \int_N^M \frac{S( u ) }{u^{3/2 } } \mathrm d u .$$ As $M\to \infty$  this converges because   the integral converges absolutely since it is $$
 O\left(  \int_N^M  \frac{ 1 }{u^{3/2 } } \mathrm d u \right)=
 O\left(  \frac{2}{N^{1/2}} -  \frac{2}{M^{1/2}} \right)=
  O(N^{-1/2})
  .$$ Hence, the limit satisfies $$\sum_{n=N+1 }^\infty
\chi(n) n^{-1/2} = 
O\left( \frac{1 }{M^{1/2}} \right)
-\frac{S(N)}{N^{1/2}}
  O(N^{-1/2})
  ,$$ which, by $M >N$  and the estimate $|S(u)| =O(1)$ is $  O(N^{-1/2})
$.
