I am self studying analytic number theory from class notes of a senior and in it I am unable to deduce an inequality which is not proved .

Assume $b(m) = \sum_{s, t} \frac{1} {log(s) log(t) } $with these 3 conditions:

  1. $2\leq s \leq n$
  2. $2\leq t \leq n$
  3. $s+t= m $

It is to be proved that if $m \leq n$, then $b(m) \geq (m-3) log^{-2} n $

Please help.

  • $\begingroup$ If $m\le n$, then $n$ should not play any role. $\endgroup$ – user Apr 26 '20 at 20:21
  • $\begingroup$ @user can you please write an answer which is a bit more ellaborated . I am confused. $\endgroup$ – Tim Apr 26 '20 at 20:35
  • $\begingroup$ It is not an answer but a comment aimed on improvement of your question. Do you see that if $n\ge m$, it can be replaced with $m-2$? $\endgroup$ – user Apr 26 '20 at 20:42
  • $\begingroup$ @user m -2 is not greater than Or equal to m. So, how can you replace it? $\endgroup$ – Tim Apr 26 '20 at 20:46
  • $\begingroup$ @user also why do you want to " improve the question " ? Do you mean giving a hint for proof? $\endgroup$ – Tim Apr 26 '20 at 20:47

If $n\ge m$ the sum consists of $m-3$ terms. Since $\log x$ for $x\ge2$ is positive increasing function $$ \frac1{\log s \log (m-s)}>\frac1{\log^2 m}.$$

Therefore: $$b(m)>\frac{m-3}{\log^2m}\ge\frac{m-3}{\log^2n}.$$


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.