I am stuck at the following exercise:
Let $\chi$ be a non-principal character modulo $q$. Show that
$$\sum_{n \ge x} \frac{\chi(n)}{\sqrt{n}} = \mathcal{O}\bigg(\frac{1}{\sqrt{x}}\bigg)$$
My Attempt: Let $A:= \max_{n \in \{1,\ldots,q-1\}} \chi(n)$. Then we have
$$\bigg\lvert \sum_{n \ge x} \frac{\chi(n)}{\sqrt{n}} \bigg\rvert \le \sum_{n \ge x} \frac{\lvert\chi(n)\rvert}{\sqrt{n}} = A\cdot \sum_{n \ge x} \frac{1}{\sqrt{n}}.$$
Here I get stuck. I understand that $\sum_{n \ge x} \frac{1}{\sqrt{n}}$ is related to the harmonic series $H_n$ by
$$\sum_{n=1}^\infty \frac{1}{n} - H_{\lfloor x \rfloor} = \sum_{n \ge x} \frac{1}{\sqrt{n}} $$
and we know that
$$H_n = \gamma + log(n) + \mathcal{O}(1/n)$$
, where $\gamma$ is the Euler-Maceroni constant. Can we use this here someow?
The partial sum $$ A(x) \equiv \sum_{n \leq x} \chi(n) $$ is bounded; $A(t) \ll 1$.
Therefore, $$ \sum_{n \leq x} \chi(n)/n^{1/2} = A(x)/x^{1/2} + constant + (1/2) \int_{1}^{x}A(t)t^{-3/2} dt, $$ with the last integral convergent as $x$ goes to infinity.
Once these are known, we choose in the above as $$ \sum_{n \geq x} \chi(n)/n^{1/2} = - A(x)/x^{1/2} + (1/2) \int_{x}^{\infty}A(t)t^{-3/2} dt \ll x^{-1/2}. $$