# Show that $\sum_{n \ge x} \frac{\chi(n)}{\sqrt{n}} = \mathcal{O}\bigg(\frac{1}{\sqrt{x}}\bigg)$

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?

• 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. – Sangchul Lee Jun 16 '20 at 18:10
• @SangchulLee: Not $2\sqrt{x}$ but $\infty$ (which doesn't affect the rest). – metamorphy Jun 17 '20 at 8:53
• Trivially bounding by $1$, is not recommended. Instead, proceed with the partial summation. – Sungjin Kim Jun 17 '20 at 9:33
• @metamorphy, You are right. I guess I had not enough caffeine to fool myself at that moment. Thank you for pointing that out! – Sangchul Lee Jun 17 '20 at 16:36

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 $$\sum_{n \geq x} \chi(n)/n^{1/2} = - A(x)/x^{1/2} + (1/2) \int_{x}^{\infty}A(t)t^{-3/2} dt \ll x^{-1/2}.$$