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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?

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    $\begingroup$ First, your first equation does not hold in general. I guess what you intended there are inequalities $\leq$. Second, $$\sum_{n \geq x}\frac{1}{\sqrt{n}} \approx \int_{x}^{\infty}\frac{\mathrm{d}y}{\sqrt{y}} = 2\sqrt{x}.$$ So your approach will not succeed. For the proof, the idea is that one may take advantage of the oscillatory nature of the non-principal characters. For this, write $X(n) = \sum_{k=1}^{n} \chi(k)$ and note that $X(n)$ is bounded. So you may apply the summation by parts to conclude. $\endgroup$ – Sangchul Lee Jun 16 '20 at 18:10
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    $\begingroup$ @SangchulLee: Not $2\sqrt{x}$ but $\infty$ (which doesn't affect the rest). $\endgroup$ – metamorphy Jun 17 '20 at 8:53
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    $\begingroup$ Trivially bounding by $1$, is not recommended. Instead, proceed with the partial summation. $\endgroup$ – Sungjin Kim Jun 17 '20 at 9:33
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    $\begingroup$ @metamorphy, You are right. I guess I had not enough caffeine to fool myself at that moment. Thank you for pointing that out! $\endgroup$ – Sangchul Lee Jun 17 '20 at 16:36
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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 $$ \sum_{n \geq x} \chi(n)/n^{1/2} = - A(x)/x^{1/2} + (1/2) \int_{x}^{\infty}A(t)t^{-3/2} dt \ll x^{-1/2}. $$

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