Bounding a Mobius Fractional Sum












2












$begingroup$


How does one show the following estimate:



$displaystyle bigglvert 1+sum_{n leq x} mu(n) Big{frac{x}{n}Big} biggrvert leq x$, where $xgeq1$ ?



My attempt was to use the triangle inequality so that



$displaystyle bigglvert 1+sum_{n leq x} mu(n) Big{frac{x}{n}Big} biggrvert leq 1 + sum_{n leq x} Big{frac{x}{n}Big} = 1 + {x} + sum_{2leq n leq x-1} Big{frac{x}{n}Big}$,



and to try and bound the re-indexed sum by $[x]-1$ (here, $[x]$ denotes the floor of x), which will then give $1+{x}+[x]-1={x}+[x]=x$. Dividing by x would then yield the desired result, but I'm having difficulty in finding why this bound makes sense. Am I missing something obvious here? Perhaps there's an easier way to derive this? I also suspect that fiddling around with the triangle inequality is unnecessary, and this could be shown directly without much effort, but I'm just not seeing it. Any explanations would be appreciated!






number-theory summation analytic-number-theory







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    2












    $begingroup$


    How does one show the following estimate:



    $displaystyle bigglvert 1+sum_{n leq x} mu(n) Big{frac{x}{n}Big} biggrvert leq x$, where $xgeq1$ ?



    My attempt was to use the triangle inequality so that



    $displaystyle bigglvert 1+sum_{n leq x} mu(n) Big{frac{x}{n}Big} biggrvert leq 1 + sum_{n leq x} Big{frac{x}{n}Big} = 1 + {x} + sum_{2leq n leq x-1} Big{frac{x}{n}Big}$,



    and to try and bound the re-indexed sum by $[x]-1$ (here, $[x]$ denotes the floor of x), which will then give $1+{x}+[x]-1={x}+[x]=x$. Dividing by x would then yield the desired result, but I'm having difficulty in finding why this bound makes sense. Am I missing something obvious here? Perhaps there's an easier way to derive this? I also suspect that fiddling around with the triangle inequality is unnecessary, and this could be shown directly without much effort, but I'm just not seeing it. Any explanations would be appreciated!






    number-theory summation analytic-number-theory







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      How does one show the following estimate:



      $displaystyle bigglvert 1+sum_{n leq x} mu(n) Big{frac{x}{n}Big} biggrvert leq x$, where $xgeq1$ ?



      My attempt was to use the triangle inequality so that



      $displaystyle bigglvert 1+sum_{n leq x} mu(n) Big{frac{x}{n}Big} biggrvert leq 1 + sum_{n leq x} Big{frac{x}{n}Big} = 1 + {x} + sum_{2leq n leq x-1} Big{frac{x}{n}Big}$,



      and to try and bound the re-indexed sum by $[x]-1$ (here, $[x]$ denotes the floor of x), which will then give $1+{x}+[x]-1={x}+[x]=x$. Dividing by x would then yield the desired result, but I'm having difficulty in finding why this bound makes sense. Am I missing something obvious here? Perhaps there's an easier way to derive this? I also suspect that fiddling around with the triangle inequality is unnecessary, and this could be shown directly without much effort, but I'm just not seeing it. Any explanations would be appreciated!






      number-theory summation analytic-number-theory







      share|cite|improve this question









      $endgroup$




      How does one show the following estimate:



      $displaystyle bigglvert 1+sum_{n leq x} mu(n) Big{frac{x}{n}Big} biggrvert leq x$, where $xgeq1$ ?



      My attempt was to use the triangle inequality so that



      $displaystyle bigglvert 1+sum_{n leq x} mu(n) Big{frac{x}{n}Big} biggrvert leq 1 + sum_{n leq x} Big{frac{x}{n}Big} = 1 + {x} + sum_{2leq n leq x-1} Big{frac{x}{n}Big}$,



      and to try and bound the re-indexed sum by $[x]-1$ (here, $[x]$ denotes the floor of x), which will then give $1+{x}+[x]-1={x}+[x]=x$. Dividing by x would then yield the desired result, but I'm having difficulty in finding why this bound makes sense. Am I missing something obvious here? Perhaps there's an easier way to derive this? I also suspect that fiddling around with the triangle inequality is unnecessary, and this could be shown directly without much effort, but I'm just not seeing it. Any explanations would be appreciated!






      number-theory summation analytic-number-theory




      number-theory summation analytic-number-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Feb 2 at 1:23









      ButtforButtfor

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          First, I want to point out that $$sum_{nleq x}left{frac nxright}neq{x}+sum_{2leq nleq x-1}left{frac nxright}.$$ The problem is that $x$ need not be an integer (in particular, you may not have a case $n=x$, so $x-1$ might miss an integer). Now, the fractional part of a number is always bounded above by $1$, so you in fact have $$sum_{2leq nleq x}left{frac xnright}leqsum_{2leq nleq x}1=lfloor xrfloor-1.$$ Combining the steps as you have, you end up with $$left|1+sum_{nleq x}mu(n)left{frac xnright}right|leq 1+{x}+lfloor xrfloor-1=x.$$






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            Thank you very much!
            $endgroup$
            – Buttfor
            Feb 2 at 13:49












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          $begingroup$

          First, I want to point out that $$sum_{nleq x}left{frac nxright}neq{x}+sum_{2leq nleq x-1}left{frac nxright}.$$ The problem is that $x$ need not be an integer (in particular, you may not have a case $n=x$, so $x-1$ might miss an integer). Now, the fractional part of a number is always bounded above by $1$, so you in fact have $$sum_{2leq nleq x}left{frac xnright}leqsum_{2leq nleq x}1=lfloor xrfloor-1.$$ Combining the steps as you have, you end up with $$left|1+sum_{nleq x}mu(n)left{frac xnright}right|leq 1+{x}+lfloor xrfloor-1=x.$$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Thank you very much!
            $endgroup$
            – Buttfor
            Feb 2 at 13:49
















          2












          $begingroup$

          First, I want to point out that $$sum_{nleq x}left{frac nxright}neq{x}+sum_{2leq nleq x-1}left{frac nxright}.$$ The problem is that $x$ need not be an integer (in particular, you may not have a case $n=x$, so $x-1$ might miss an integer). Now, the fractional part of a number is always bounded above by $1$, so you in fact have $$sum_{2leq nleq x}left{frac xnright}leqsum_{2leq nleq x}1=lfloor xrfloor-1.$$ Combining the steps as you have, you end up with $$left|1+sum_{nleq x}mu(n)left{frac xnright}right|leq 1+{x}+lfloor xrfloor-1=x.$$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Thank you very much!
            $endgroup$
            – Buttfor
            Feb 2 at 13:49














          2












          2








          2





          $begingroup$

          First, I want to point out that $$sum_{nleq x}left{frac nxright}neq{x}+sum_{2leq nleq x-1}left{frac nxright}.$$ The problem is that $x$ need not be an integer (in particular, you may not have a case $n=x$, so $x-1$ might miss an integer). Now, the fractional part of a number is always bounded above by $1$, so you in fact have $$sum_{2leq nleq x}left{frac xnright}leqsum_{2leq nleq x}1=lfloor xrfloor-1.$$ Combining the steps as you have, you end up with $$left|1+sum_{nleq x}mu(n)left{frac xnright}right|leq 1+{x}+lfloor xrfloor-1=x.$$






          share|cite|improve this answer











          $endgroup$



          First, I want to point out that $$sum_{nleq x}left{frac nxright}neq{x}+sum_{2leq nleq x-1}left{frac nxright}.$$ The problem is that $x$ need not be an integer (in particular, you may not have a case $n=x$, so $x-1$ might miss an integer). Now, the fractional part of a number is always bounded above by $1$, so you in fact have $$sum_{2leq nleq x}left{frac xnright}leqsum_{2leq nleq x}1=lfloor xrfloor-1.$$ Combining the steps as you have, you end up with $$left|1+sum_{nleq x}mu(n)left{frac xnright}right|leq 1+{x}+lfloor xrfloor-1=x.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 2 at 17:16

























          answered Feb 2 at 1:59









          ClaytonClayton

          19.6k33288




          19.6k33288








          • 2




            $begingroup$
            Thank you very much!
            $endgroup$
            – Buttfor
            Feb 2 at 13:49














          • 2




            $begingroup$
            Thank you very much!
            $endgroup$
            – Buttfor
            Feb 2 at 13:49








          2




          2




          $begingroup$
          Thank you very much!
          $endgroup$
          – Buttfor
          Feb 2 at 13:49




          $begingroup$
          Thank you very much!
          $endgroup$
          – Buttfor
          Feb 2 at 13:49


















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