Bounding sum (log factor)

Question

I want to prove that $$ \sum_{\substack{1\leq n\leq T \\ n\neq m}}n^{-\frac{1}{2}}\left|\log \frac{m}{n}\right|^{-1}\ll T^{\frac{1}{2}}\log T $$ for any $1\leq m \leq T$. Do you have any hint how I should proceed? Thank you a lot!

Answer 1

I will sketch a proof: basically, you split the sum in 3 (one or even two subsums may be empty or not complete depending on $m$), namely $S_1$ for $1 \leq n \leq m-1$, $S_2$ for $m+1 \leq n \leq 2m-1$, $S_3$ for $2m \leq n \leq T$.

$S_3$ is the easiest to deal with as there $\left|\log \frac{m}{n}\right| \geq \log 2$, so $S_3 \leq \frac{1}{\log 2}\sum_{\substack{2m\leq n\leq T}}n^{-\frac{1}{2}} = O(T^{\frac{1}{2}})$ with an absolute implied constant

$S_1$ and $S_2$ are similar as in each you use the change of variables $n=m-k, 1 \leq k \leq m-1$ and $n=m+k, 1 \leq k \leq m-1$ respectively as well as the estimates $|\log({1 - \frac{k}{m}})| \geq \frac{k}{m}, 1 \leq k \leq m-1$, $\log({1 + \frac{k}{m}}) \geq \frac{k}{2m}, 1 \leq k \leq m-1$ which are well known (and trivial to derive from the properties of $\log(1-x), \log(1+x), 0 \leq x < 1$

$S_2 \leq 2m\sum_{\substack{1\leq k\leq m-1}}(m+k)^{-\frac{1}{2}}k^{-1} \leq 2m^{\frac{1}{2}}\sum_{\substack{1\leq k\leq m-1}}k^{-1} \leq 2m^{\frac{1}{2}}\log m \leq 2T^{\frac{1}{2}}\log T$, since $(m+k)^{\frac{1}{2}} \geq m^{\frac{1}{2}}, m \leq T$ (and obviously the estimate works if the sum is not complete when $2m-1 > T$)

$S_1 \leq m\sum_{\substack{1\leq k\leq m-1}}(m-k)^{-\frac{1}{2}}k^{-1} =m\sum_{\substack{1\leq k\leq m-1}}(m-k)^{\frac{1}{2}}(m-k)^{-1}k^{-1} = \sum_{\substack{1\leq k\leq m-1}}(m-k)^{\frac{1}{2}}((m-k)^{-1}+k^{-1}) \leq \sqrt{m}\sum_{\substack{1\leq k\leq m-1}}((m-k)^{-1}+k^{-1}) \leq 2\sqrt{m}\log m \leq 2T^{\frac{1}{2}}\log T$