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This problem is from Apostol's Analytic Number Theory, Chapter $9$.

The problem:

Let $\chi$ be a primitive Dirichlet character mod $k$. Prove that if $N< M$ we have $$ \bigg|\sum_{m=N+1}^M\dfrac{\chi(m)}{m}\bigg|< \dfrac{2}{N+1}\sqrt{k}\log k$$

So I'm trying to use the Pólya Inequality which states that if $k\geq 3$ and $\chi$ is primitive character mod $k$, then : $$\bigg|\sum_{m\leq x}\chi(m)\bigg|< \sqrt{k}\log k.$$

And I have used the Abel's Summation formula, I let $G(x)=\sum_{m\leq x}\chi(m)$ and arrived at this weaker estimate: \begin{align*} \sum_{m=N+1}^M \frac{\chi(m)}{m}&=\sum_{N< m\leq M}\frac{\chi(m)}{m}\\ &=\frac{G(M)}{M}-\frac{G(N)}{N}+\int_N^M \frac{G(t)}{t^2}dt\\ \therefore \bigg|\sum_{m=N+1}^M\frac{\chi(m)}{m}\bigg|&\leq \frac{|G(M)|}{M}+\frac{|G(N)|}{N}+\int_N^M\frac{|G(t)|}{t^2}dt\\ &< \bigg[\frac1M+\frac1N\bigg]\sqrt{k}\log k+\sqrt{k}\log k\bigg[\frac1N-\frac1M\bigg]\\ &=\frac2N\sqrt{k}\log k \end{align*}

I think I have chosen a right road because I can tell where is the numerator $2$ and the denominator $N$ comes, but I still miss why the question ask me to prove where the denominator is $N+1$.

Any completely different solution will also be appreciated.

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    $\begingroup$ "weaker" estimate because you obtain $2/N$ instead of $2/(N+1)$ ? You shouldn't care about that. Partial summation and Abel summation are quite the same thing, they give quite the same result. And the estimate for $\sup_N |\sum_{n=1}^N \chi(n)|$ is obtained using that the discrete Fourier transform of $\chi$ primitive is $\overline{\chi(n)} G(\chi),|G(\chi)| = \sqrt{k}$ $\endgroup$ – reuns Mar 5 at 1:41
  • $\begingroup$ @reuns But we cannot say that $2/N< 2/(N+1)$ is it? I want to find out how people can get $2/(N+1)$ instead of $2/N$. $\endgroup$ – kelvin hong 方 Mar 6 at 2:03
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I have got the answer, the original inequality is indeed true. Here is the solution:

Let $S(x)=\sum_{n\leq x}\chi(n)$, then \begin{align*}\sum_{m=N+1}^M\frac{\chi(m)}{m}&=\sum_{m=N+1}^M\frac{S(m)-S(m-1)}{m}\\ &=\sum_{m=N+1}^M\frac{S(m)}{m}-\sum_{m=N}^{M-1}\frac{S(m)}{m+1}\\ &=\frac{S(M)}{M}-\frac{S(N)}{N+1}+\sum_{m=N+1}^{M-1}S(m)\bigg[\frac{1}{m}-\frac1{m+1}\bigg]\\\therefore \bigg|\sum_{m=N+1}^M\frac{\chi(m)}m\bigg|&\leq \frac{|S(M)|}{M}+\frac{|S(N)|}{N+1}+\sum_{m=N+1}^{M-1}|S(m)|\bigg[\frac1m-\frac1{m+1}\bigg]\\ &< \sqrt{k}\log k\bigg[\frac1{M}+\frac1{N+1}+\sum_{m=N+1}^{M-1}\bigg(\frac1m-\frac1{m+1}\bigg)\bigg]\\ &=\sqrt{k}\log k\bigg[\frac1M+\frac1{N+1}+\bigg(\frac1{N+1}-\frac1{N+2}+\cdots+\frac1{M-1}-\frac1M\bigg)\bigg]\\ &=\frac2{N+1}\sqrt k\log k.\end{align*}

I don't know is it because Abel's Summation is too general so that it cannot hold the stronger condition. I think the sum of the form $\sum_{m\leq x}a(n)f(n)$ can be improved when $f(n)$ is a decreasing function.

Thanks for helping.

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$\left|\sum_{m=N+1}^{M} \frac{\chi(m)}{m}\right| \leq \frac{1}{N+1} \bigg|\sum_{m=N+1}^{M} \chi(m) \bigg| \leq \frac{1}{N+1}\bigg(\bigg|\sum_{m=1}^{M} \chi(m) \bigg|+\bigg|\sum_{m=1}^{N} \chi(m) \bigg| \bigg) < \frac{2}{N+1} \sqrt{k} \log(k)$

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    $\begingroup$ You cannot simply take out the denominator $N+1$ because $\chi(m)$ is not guarantee as non-negative. It is even complex. $\endgroup$ – kelvin hong 方 Mar 2 at 10:23

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