limit of implicit sequences












3














In a HS problem, we study the functions



$$f_n:x longmapsto (x-n)ln x - xln(x-n)$$ for $n$ natural with $nge 5$



First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $alpha_n$ that satisifes $n+1 < alpha_n <n+2$.



From last inequality it's clear that $(alpha_n)$ diverges to $+infty$.



The last question asks to prove that $limlimits_{n to +infty} (alpha_n - n)=1$ and then to evaluate $limlimits_{n to +infty} (alpha_{n+1} - alpha_n)$.



I'm stuck with this last question.






calculus sequences-and-series







share|cite|improve this question





























    3














    In a HS problem, we study the functions



    $$f_n:x longmapsto (x-n)ln x - xln(x-n)$$ for $n$ natural with $nge 5$



    First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $alpha_n$ that satisifes $n+1 < alpha_n <n+2$.



    From last inequality it's clear that $(alpha_n)$ diverges to $+infty$.



    The last question asks to prove that $limlimits_{n to +infty} (alpha_n - n)=1$ and then to evaluate $limlimits_{n to +infty} (alpha_{n+1} - alpha_n)$.



    I'm stuck with this last question.






    calculus sequences-and-series







    share|cite|improve this question



























      3












      3








      3







      In a HS problem, we study the functions



      $$f_n:x longmapsto (x-n)ln x - xln(x-n)$$ for $n$ natural with $nge 5$



      First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $alpha_n$ that satisifes $n+1 < alpha_n <n+2$.



      From last inequality it's clear that $(alpha_n)$ diverges to $+infty$.



      The last question asks to prove that $limlimits_{n to +infty} (alpha_n - n)=1$ and then to evaluate $limlimits_{n to +infty} (alpha_{n+1} - alpha_n)$.



      I'm stuck with this last question.






      calculus sequences-and-series







      share|cite|improve this question















      In a HS problem, we study the functions



      $$f_n:x longmapsto (x-n)ln x - xln(x-n)$$ for $n$ natural with $nge 5$



      First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $alpha_n$ that satisifes $n+1 < alpha_n <n+2$.



      From last inequality it's clear that $(alpha_n)$ diverges to $+infty$.



      The last question asks to prove that $limlimits_{n to +infty} (alpha_n - n)=1$ and then to evaluate $limlimits_{n to +infty} (alpha_{n+1} - alpha_n)$.



      I'm stuck with this last question.






      calculus sequences-and-series




      calculus sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 51 mins ago

























      asked 2 hours ago









      Oussama Sarih

      46827




      46827






















          2 Answers
          2






          active

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          2














          For the first part note that



          $$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$



          Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.



          For the second part



          $$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$






          share|cite|improve this answer





























            2














            This could be a stupid answer.



            You look for the zero of
            $$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
            $$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
            $$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$






            share|cite|improve this answer





















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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              2














              For the first part note that



              $$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$



              Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.



              For the second part



              $$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$






              share|cite|improve this answer


























                2














                For the first part note that



                $$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$



                Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.



                For the second part



                $$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$






                share|cite|improve this answer
























                  2












                  2








                  2






                  For the first part note that



                  $$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$



                  Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.



                  For the second part



                  $$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$






                  share|cite|improve this answer












                  For the first part note that



                  $$frac{alpha_n-n}{log(alpha_n - n)} = frac{alpha_n}{log(alpha_n)} $$



                  Since $alpha_n to infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < alpha_n - n < 2$ for all $n$, this implies that $log(alpha_n-n) to 0$ i.e. $alpha_n - n to 1$.



                  For the second part



                  $$ begin{align} lim(alpha_{n+1} - alpha_n) & = lim((a_{n+1} - (n+1)) - (alpha_n - n) + 1) \ & = lim(alpha_{n+1} - (n+1)) - lim(alpha_n - n) + 1 \ & = 1 - 1 + 1 \ & = 1 end{align}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  ODF

                  1,096410




                  1,096410























                      2














                      This could be a stupid answer.



                      You look for the zero of
                      $$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
                      $$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
                      $$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$






                      share|cite|improve this answer


























                        2














                        This could be a stupid answer.



                        You look for the zero of
                        $$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
                        $$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
                        $$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$






                        share|cite|improve this answer
























                          2












                          2








                          2






                          This could be a stupid answer.



                          You look for the zero of
                          $$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
                          $$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
                          $$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$






                          share|cite|improve this answer












                          This could be a stupid answer.



                          You look for the zero of
                          $$f_n=(x-n)log( x) - xlog(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+epsilon$ to make
                          $$f_n=(1+epsilon ) log (n+1+epsilon)-(n+1+epsilon) log (1+epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
                          $$f_n=-n log (1+epsilon)-(1+epsilon)log left(frac{1+epsilon }{n}right)+Oleft(frac{1}{n}right)$$ So, if $nto infty$, $epsilonto 0$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          Claude Leibovici

                          118k1157132




                          118k1157132






























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