Taylor series of $F(n) = 1/(n-1)^2 - 1/n^2$ around large n?












0












$begingroup$


I am lost on this taylor series. The hint says to let x = 1/n and expand around x = 0, but I can't make any progress.



I am also confused why this hint is helpful. Can't I just expand around large n? Then I can assume 1/n^2, 1/n^3... and the other higher order derivatives to be about 0, hence the taylor series expansion would be 0. Any advice?






calculus taylor-expansion







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    0












    $begingroup$


    I am lost on this taylor series. The hint says to let x = 1/n and expand around x = 0, but I can't make any progress.



    I am also confused why this hint is helpful. Can't I just expand around large n? Then I can assume 1/n^2, 1/n^3... and the other higher order derivatives to be about 0, hence the taylor series expansion would be 0. Any advice?






    calculus taylor-expansion







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am lost on this taylor series. The hint says to let x = 1/n and expand around x = 0, but I can't make any progress.



      I am also confused why this hint is helpful. Can't I just expand around large n? Then I can assume 1/n^2, 1/n^3... and the other higher order derivatives to be about 0, hence the taylor series expansion would be 0. Any advice?






      calculus taylor-expansion







      share|cite|improve this question









      $endgroup$




      I am lost on this taylor series. The hint says to let x = 1/n and expand around x = 0, but I can't make any progress.



      I am also confused why this hint is helpful. Can't I just expand around large n? Then I can assume 1/n^2, 1/n^3... and the other higher order derivatives to be about 0, hence the taylor series expansion would be 0. Any advice?






      calculus taylor-expansion




      calculus taylor-expansion






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      asked Jan 11 at 4:09









      GoldnameGoldname

      1,475930




      1,475930






















          2 Answers
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          1












          $begingroup$

          $begin{array}\
          F(n)
          &= dfrac1{(n-1)^2} - dfrac1{n^2}\
          &= dfrac1{n^2}left(dfrac1{(1-1/n)^2} -1right)\
          &= displaystyledfrac1{n^2}left(sum_{k=0}^{infty}dfrac{k+1}{n^{2k}} -1right)\
          &= displaystyledfrac1{n^2}sum_{k=1}^{infty}dfrac{k+1}{n^{2k}} \
          &= displaystylesum_{k=1}^{infty}dfrac{k+1}{n^{2k+2}} \
          &= displaystylesum_{k=2}^{infty}dfrac{k}{n^{2k}} \
          end{array}
          $






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you know why wolfram alpha ends the term with O(x^9)? wolframalpha.com/input/?i=taylor+series+of+x%5E2*(1%2F(1-x)%5E2-1)+at+x+%3D+0
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            Wolfy works in mysterious ways its wonders to perform.
            $endgroup$
            – marty cohen
            Jan 12 at 6:03



















          0












          $begingroup$

          Let us do it with $x=frac 1n$
          $$F(x)=frac{(2-x) x^3}{(1-x)^2}$$ Now, use the binomial expansion for the denominator
          $$frac{1}{(1-x)^2}=1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdots$$
          $$F(x)=(2-x) x^3 left( 1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdotsright)$$
          $$F(x)=2 x^3+3 x^4+4 x^5+5 x^6+6 x^7+7 x^8+8 x^9+cdots$$
          $$F(x)=sum_{k=2}^infty k x^{k+1 }$$
          $$F(n)=sum_{k=2}^infty frac k {n^{k+1 }}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see thanks, but wolfram alpha ends with O(x^9) as the ending term, which doesn't agree with anything
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            @Goldname. Just because the truncated to $O(x^9)$ that is to say that they consider the terms up to $x^8$. It is more interesting to have the infinie sum (just as marty cohen also did). Truncate whereever you wish.
            $endgroup$
            – Claude Leibovici
            Jan 12 at 4:17












          Your Answer





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          2 Answers
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          1












          $begingroup$

          $begin{array}\
          F(n)
          &= dfrac1{(n-1)^2} - dfrac1{n^2}\
          &= dfrac1{n^2}left(dfrac1{(1-1/n)^2} -1right)\
          &= displaystyledfrac1{n^2}left(sum_{k=0}^{infty}dfrac{k+1}{n^{2k}} -1right)\
          &= displaystyledfrac1{n^2}sum_{k=1}^{infty}dfrac{k+1}{n^{2k}} \
          &= displaystylesum_{k=1}^{infty}dfrac{k+1}{n^{2k+2}} \
          &= displaystylesum_{k=2}^{infty}dfrac{k}{n^{2k}} \
          end{array}
          $






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you know why wolfram alpha ends the term with O(x^9)? wolframalpha.com/input/?i=taylor+series+of+x%5E2*(1%2F(1-x)%5E2-1)+at+x+%3D+0
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            Wolfy works in mysterious ways its wonders to perform.
            $endgroup$
            – marty cohen
            Jan 12 at 6:03
















          1












          $begingroup$

          $begin{array}\
          F(n)
          &= dfrac1{(n-1)^2} - dfrac1{n^2}\
          &= dfrac1{n^2}left(dfrac1{(1-1/n)^2} -1right)\
          &= displaystyledfrac1{n^2}left(sum_{k=0}^{infty}dfrac{k+1}{n^{2k}} -1right)\
          &= displaystyledfrac1{n^2}sum_{k=1}^{infty}dfrac{k+1}{n^{2k}} \
          &= displaystylesum_{k=1}^{infty}dfrac{k+1}{n^{2k+2}} \
          &= displaystylesum_{k=2}^{infty}dfrac{k}{n^{2k}} \
          end{array}
          $






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you know why wolfram alpha ends the term with O(x^9)? wolframalpha.com/input/?i=taylor+series+of+x%5E2*(1%2F(1-x)%5E2-1)+at+x+%3D+0
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            Wolfy works in mysterious ways its wonders to perform.
            $endgroup$
            – marty cohen
            Jan 12 at 6:03














          1












          1








          1





          $begingroup$

          $begin{array}\
          F(n)
          &= dfrac1{(n-1)^2} - dfrac1{n^2}\
          &= dfrac1{n^2}left(dfrac1{(1-1/n)^2} -1right)\
          &= displaystyledfrac1{n^2}left(sum_{k=0}^{infty}dfrac{k+1}{n^{2k}} -1right)\
          &= displaystyledfrac1{n^2}sum_{k=1}^{infty}dfrac{k+1}{n^{2k}} \
          &= displaystylesum_{k=1}^{infty}dfrac{k+1}{n^{2k+2}} \
          &= displaystylesum_{k=2}^{infty}dfrac{k}{n^{2k}} \
          end{array}
          $






          share|cite|improve this answer











          $endgroup$



          $begin{array}\
          F(n)
          &= dfrac1{(n-1)^2} - dfrac1{n^2}\
          &= dfrac1{n^2}left(dfrac1{(1-1/n)^2} -1right)\
          &= displaystyledfrac1{n^2}left(sum_{k=0}^{infty}dfrac{k+1}{n^{2k}} -1right)\
          &= displaystyledfrac1{n^2}sum_{k=1}^{infty}dfrac{k+1}{n^{2k}} \
          &= displaystylesum_{k=1}^{infty}dfrac{k+1}{n^{2k+2}} \
          &= displaystylesum_{k=2}^{infty}dfrac{k}{n^{2k}} \
          end{array}
          $







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 0:05









          Simply Beautiful Art

          50.9k580186




          50.9k580186










          answered Jan 11 at 4:20









          marty cohenmarty cohen

          75.2k549130




          75.2k549130












          • $begingroup$
            Do you know why wolfram alpha ends the term with O(x^9)? wolframalpha.com/input/?i=taylor+series+of+x%5E2*(1%2F(1-x)%5E2-1)+at+x+%3D+0
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            Wolfy works in mysterious ways its wonders to perform.
            $endgroup$
            – marty cohen
            Jan 12 at 6:03


















          • $begingroup$
            Do you know why wolfram alpha ends the term with O(x^9)? wolframalpha.com/input/?i=taylor+series+of+x%5E2*(1%2F(1-x)%5E2-1)+at+x+%3D+0
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            Wolfy works in mysterious ways its wonders to perform.
            $endgroup$
            – marty cohen
            Jan 12 at 6:03
















          $begingroup$
          Do you know why wolfram alpha ends the term with O(x^9)? wolframalpha.com/input/?i=taylor+series+of+x%5E2*(1%2F(1-x)%5E2-1)+at+x+%3D+0
          $endgroup$
          – Goldname
          Jan 12 at 0:00




          $begingroup$
          Do you know why wolfram alpha ends the term with O(x^9)? wolframalpha.com/input/?i=taylor+series+of+x%5E2*(1%2F(1-x)%5E2-1)+at+x+%3D+0
          $endgroup$
          – Goldname
          Jan 12 at 0:00












          $begingroup$
          Wolfy works in mysterious ways its wonders to perform.
          $endgroup$
          – marty cohen
          Jan 12 at 6:03




          $begingroup$
          Wolfy works in mysterious ways its wonders to perform.
          $endgroup$
          – marty cohen
          Jan 12 at 6:03











          0












          $begingroup$

          Let us do it with $x=frac 1n$
          $$F(x)=frac{(2-x) x^3}{(1-x)^2}$$ Now, use the binomial expansion for the denominator
          $$frac{1}{(1-x)^2}=1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdots$$
          $$F(x)=(2-x) x^3 left( 1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdotsright)$$
          $$F(x)=2 x^3+3 x^4+4 x^5+5 x^6+6 x^7+7 x^8+8 x^9+cdots$$
          $$F(x)=sum_{k=2}^infty k x^{k+1 }$$
          $$F(n)=sum_{k=2}^infty frac k {n^{k+1 }}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see thanks, but wolfram alpha ends with O(x^9) as the ending term, which doesn't agree with anything
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            @Goldname. Just because the truncated to $O(x^9)$ that is to say that they consider the terms up to $x^8$. It is more interesting to have the infinie sum (just as marty cohen also did). Truncate whereever you wish.
            $endgroup$
            – Claude Leibovici
            Jan 12 at 4:17
















          0












          $begingroup$

          Let us do it with $x=frac 1n$
          $$F(x)=frac{(2-x) x^3}{(1-x)^2}$$ Now, use the binomial expansion for the denominator
          $$frac{1}{(1-x)^2}=1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdots$$
          $$F(x)=(2-x) x^3 left( 1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdotsright)$$
          $$F(x)=2 x^3+3 x^4+4 x^5+5 x^6+6 x^7+7 x^8+8 x^9+cdots$$
          $$F(x)=sum_{k=2}^infty k x^{k+1 }$$
          $$F(n)=sum_{k=2}^infty frac k {n^{k+1 }}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see thanks, but wolfram alpha ends with O(x^9) as the ending term, which doesn't agree with anything
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            @Goldname. Just because the truncated to $O(x^9)$ that is to say that they consider the terms up to $x^8$. It is more interesting to have the infinie sum (just as marty cohen also did). Truncate whereever you wish.
            $endgroup$
            – Claude Leibovici
            Jan 12 at 4:17














          0












          0








          0





          $begingroup$

          Let us do it with $x=frac 1n$
          $$F(x)=frac{(2-x) x^3}{(1-x)^2}$$ Now, use the binomial expansion for the denominator
          $$frac{1}{(1-x)^2}=1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdots$$
          $$F(x)=(2-x) x^3 left( 1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdotsright)$$
          $$F(x)=2 x^3+3 x^4+4 x^5+5 x^6+6 x^7+7 x^8+8 x^9+cdots$$
          $$F(x)=sum_{k=2}^infty k x^{k+1 }$$
          $$F(n)=sum_{k=2}^infty frac k {n^{k+1 }}$$






          share|cite|improve this answer









          $endgroup$



          Let us do it with $x=frac 1n$
          $$F(x)=frac{(2-x) x^3}{(1-x)^2}$$ Now, use the binomial expansion for the denominator
          $$frac{1}{(1-x)^2}=1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdots$$
          $$F(x)=(2-x) x^3 left( 1+2 x+3 x^2+4 x^3+5 x^4+6 x^5+7 x^6+cdotsright)$$
          $$F(x)=2 x^3+3 x^4+4 x^5+5 x^6+6 x^7+7 x^8+8 x^9+cdots$$
          $$F(x)=sum_{k=2}^infty k x^{k+1 }$$
          $$F(n)=sum_{k=2}^infty frac k {n^{k+1 }}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 4:52









          Claude LeiboviciClaude Leibovici

          125k1158135




          125k1158135












          • $begingroup$
            I see thanks, but wolfram alpha ends with O(x^9) as the ending term, which doesn't agree with anything
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            @Goldname. Just because the truncated to $O(x^9)$ that is to say that they consider the terms up to $x^8$. It is more interesting to have the infinie sum (just as marty cohen also did). Truncate whereever you wish.
            $endgroup$
            – Claude Leibovici
            Jan 12 at 4:17


















          • $begingroup$
            I see thanks, but wolfram alpha ends with O(x^9) as the ending term, which doesn't agree with anything
            $endgroup$
            – Goldname
            Jan 12 at 0:00










          • $begingroup$
            @Goldname. Just because the truncated to $O(x^9)$ that is to say that they consider the terms up to $x^8$. It is more interesting to have the infinie sum (just as marty cohen also did). Truncate whereever you wish.
            $endgroup$
            – Claude Leibovici
            Jan 12 at 4:17
















          $begingroup$
          I see thanks, but wolfram alpha ends with O(x^9) as the ending term, which doesn't agree with anything
          $endgroup$
          – Goldname
          Jan 12 at 0:00




          $begingroup$
          I see thanks, but wolfram alpha ends with O(x^9) as the ending term, which doesn't agree with anything
          $endgroup$
          – Goldname
          Jan 12 at 0:00












          $begingroup$
          @Goldname. Just because the truncated to $O(x^9)$ that is to say that they consider the terms up to $x^8$. It is more interesting to have the infinie sum (just as marty cohen also did). Truncate whereever you wish.
          $endgroup$
          – Claude Leibovici
          Jan 12 at 4:17




          $begingroup$
          @Goldname. Just because the truncated to $O(x^9)$ that is to say that they consider the terms up to $x^8$. It is more interesting to have the infinie sum (just as marty cohen also did). Truncate whereever you wish.
          $endgroup$
          – Claude Leibovici
          Jan 12 at 4:17


















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