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?

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ @SangchulLee: Not $2\sqrt{x}$ but $\infty$ (which doesn't affect the rest). $\endgroup$ – metamorphy Jun 17 '20 at 8:53
  • 1
    $\begingroup$ Trivially bounding by $1$, is not recommended. Instead, proceed with the partial summation. $\endgroup$ – Sungjin Kim Jun 17 '20 at 9:33
  • 1
    $\begingroup$ @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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.