equivalent of $frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$












1














The aim is to prove that $u_n=frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$ is equivalent to $2frac{ln n}{n}$.



By using standard estimations, my book obtains that



$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)$



I don't understand the following : they write that the previous expression is equal to



$frac{2ln n}{n} + o(frac{ln n}{n})$. Why ?






calculus







share|cite|improve this question



























    1














    The aim is to prove that $u_n=frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$ is equivalent to $2frac{ln n}{n}$.



    By using standard estimations, my book obtains that



    $u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)$



    I don't understand the following : they write that the previous expression is equal to



    $frac{2ln n}{n} + o(frac{ln n}{n})$. Why ?






    calculus







    share|cite|improve this question

























      1












      1








      1







      The aim is to prove that $u_n=frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$ is equivalent to $2frac{ln n}{n}$.



      By using standard estimations, my book obtains that



      $u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)$



      I don't understand the following : they write that the previous expression is equal to



      $frac{2ln n}{n} + o(frac{ln n}{n})$. Why ?






      calculus







      share|cite|improve this question













      The aim is to prove that $u_n=frac{(n+1)^{frac{n+1}{n}}-(n-1)^{frac{n-1}{n}}}{n}$ is equivalent to $2frac{ln n}{n}$.



      By using standard estimations, my book obtains that



      $u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)$



      I don't understand the following : they write that the previous expression is equal to



      $frac{2ln n}{n} + o(frac{ln n}{n})$. Why ?






      calculus




      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 27 '18 at 21:28









      trocho

      8519




      8519






















          1 Answer
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          1














          We have that



          $$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$



          and



          $$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$



          $$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$






          share|cite|improve this answer





















          • Thanks ! I get it.
            – trocho
            Nov 27 '18 at 21:47










          • @trocho You are welcome! Bye
            – gimusi
            Nov 27 '18 at 21:48











          Your Answer





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          1 Answer
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          active

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          active

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          1














          We have that



          $$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$



          and



          $$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$



          $$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$






          share|cite|improve this answer





















          • Thanks ! I get it.
            – trocho
            Nov 27 '18 at 21:47










          • @trocho You are welcome! Bye
            – gimusi
            Nov 27 '18 at 21:48
















          1














          We have that



          $$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$



          and



          $$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$



          $$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$






          share|cite|improve this answer





















          • Thanks ! I get it.
            – trocho
            Nov 27 '18 at 21:47










          • @trocho You are welcome! Bye
            – gimusi
            Nov 27 '18 at 21:48














          1












          1








          1






          We have that



          $$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$



          and



          $$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$



          $$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$






          share|cite|improve this answer












          We have that



          $$u_n=frac{e^{ln n}}{n}left(e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}right)=e^{frac{ln n}{n}+o(frac{ln n}{n})}- e^{-frac{ln n}{n}+o(frac{ln n}{n})}$$



          and



          $$e^{frac{ln n}{n}+o(frac{ln n}{n})}=1+frac{ln n}{n}+oleft(frac{ln n}{n}right)$$



          $$e^{-frac{ln n}{n}+o(frac{ln n}{n})}=1-frac{ln n}{n}+oleft(frac{ln n}{n}right)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 21:46









          gimusi

          1




          1












          • Thanks ! I get it.
            – trocho
            Nov 27 '18 at 21:47










          • @trocho You are welcome! Bye
            – gimusi
            Nov 27 '18 at 21:48


















          • Thanks ! I get it.
            – trocho
            Nov 27 '18 at 21:47










          • @trocho You are welcome! Bye
            – gimusi
            Nov 27 '18 at 21:48
















          Thanks ! I get it.
          – trocho
          Nov 27 '18 at 21:47




          Thanks ! I get it.
          – trocho
          Nov 27 '18 at 21:47












          @trocho You are welcome! Bye
          – gimusi
          Nov 27 '18 at 21:48




          @trocho You are welcome! Bye
          – gimusi
          Nov 27 '18 at 21:48


















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