Use Differentiability of Power Series to find the sum $F(x) = sum_n=0^infty frac(x+1)^n+1n+1$

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$$F(x) = sum_n=0^infty frac(x+1)^n+1n+1$$



I found the following values after differentiating:



$$
(x+1)^n, n(x+1)^n-1, n(n-1)(x+1)^n-2
$$



It looks a lot like the Sum for $F(x)= 1/(1-x)$, but i'm not sure where to proceed from here. Thanks for your help.




calculus sequences-and-series power-series




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    up vote
    0
    down vote

    favorite












    $$F(x) = sum_n=0^infty frac(x+1)^n+1n+1$$



    I found the following values after differentiating:



    $$
    (x+1)^n, n(x+1)^n-1, n(n-1)(x+1)^n-2
    $$



    It looks a lot like the Sum for $F(x)= 1/(1-x)$, but i'm not sure where to proceed from here. Thanks for your help.




    calculus sequences-and-series power-series




    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $$F(x) = sum_n=0^infty frac(x+1)^n+1n+1$$



      I found the following values after differentiating:



      $$
      (x+1)^n, n(x+1)^n-1, n(n-1)(x+1)^n-2
      $$



      It looks a lot like the Sum for $F(x)= 1/(1-x)$, but i'm not sure where to proceed from here. Thanks for your help.




      calculus sequences-and-series power-series




      share|cite|improve this question














      $$F(x) = sum_n=0^infty frac(x+1)^n+1n+1$$



      I found the following values after differentiating:



      $$
      (x+1)^n, n(x+1)^n-1, n(n-1)(x+1)^n-2
      $$



      It looks a lot like the Sum for $F(x)= 1/(1-x)$, but i'm not sure where to proceed from here. Thanks for your help.





      calculus sequences-and-series power-series





      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 19 at 4:33









      AVK

      1,7241515




      1,7241515










      asked Aug 19 at 4:20









      AJGK

      306




      306




















          1 Answer
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          With differentiating of
          $$f(x)=sum_ngeq0dfrac(x+1)^n+1n+1$$
          we have
          $$f'(x)=sum_ngeq0(x+1)^n=dfrac11-(x+1)=-dfrac1x$$
          valid for $|x+1|<1$, then with integration of both sides
          $$f(x)=int_-1^xdfrac1-t dt=-ln(-x)$$






          share|cite|improve this answer






















          • Since I found the radius of convergence to be [-2, 0), would I then plug in (-2) such as f(x) = -ln(-(-2)) = -ln(2) ?
            – AJGK
            Aug 19 at 5:39










          • Of course, it is.
            – Nosrati
            Aug 19 at 5:45










          Your Answer




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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          With differentiating of
          $$f(x)=sum_ngeq0dfrac(x+1)^n+1n+1$$
          we have
          $$f'(x)=sum_ngeq0(x+1)^n=dfrac11-(x+1)=-dfrac1x$$
          valid for $|x+1|<1$, then with integration of both sides
          $$f(x)=int_-1^xdfrac1-t dt=-ln(-x)$$






          share|cite|improve this answer






















          • Since I found the radius of convergence to be [-2, 0), would I then plug in (-2) such as f(x) = -ln(-(-2)) = -ln(2) ?
            – AJGK
            Aug 19 at 5:39










          • Of course, it is.
            – Nosrati
            Aug 19 at 5:45














          up vote
          2
          down vote



          accepted










          With differentiating of
          $$f(x)=sum_ngeq0dfrac(x+1)^n+1n+1$$
          we have
          $$f'(x)=sum_ngeq0(x+1)^n=dfrac11-(x+1)=-dfrac1x$$
          valid for $|x+1|<1$, then with integration of both sides
          $$f(x)=int_-1^xdfrac1-t dt=-ln(-x)$$






          share|cite|improve this answer






















          • Since I found the radius of convergence to be [-2, 0), would I then plug in (-2) such as f(x) = -ln(-(-2)) = -ln(2) ?
            – AJGK
            Aug 19 at 5:39










          • Of course, it is.
            – Nosrati
            Aug 19 at 5:45












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          With differentiating of
          $$f(x)=sum_ngeq0dfrac(x+1)^n+1n+1$$
          we have
          $$f'(x)=sum_ngeq0(x+1)^n=dfrac11-(x+1)=-dfrac1x$$
          valid for $|x+1|<1$, then with integration of both sides
          $$f(x)=int_-1^xdfrac1-t dt=-ln(-x)$$






          share|cite|improve this answer














          With differentiating of
          $$f(x)=sum_ngeq0dfrac(x+1)^n+1n+1$$
          we have
          $$f'(x)=sum_ngeq0(x+1)^n=dfrac11-(x+1)=-dfrac1x$$
          valid for $|x+1|<1$, then with integration of both sides
          $$f(x)=int_-1^xdfrac1-t dt=-ln(-x)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 19 at 4:49

























          answered Aug 19 at 4:35









          Nosrati

          20.7k41644




          20.7k41644











          • Since I found the radius of convergence to be [-2, 0), would I then plug in (-2) such as f(x) = -ln(-(-2)) = -ln(2) ?
            – AJGK
            Aug 19 at 5:39










          • Of course, it is.
            – Nosrati
            Aug 19 at 5:45
















          • Since I found the radius of convergence to be [-2, 0), would I then plug in (-2) such as f(x) = -ln(-(-2)) = -ln(2) ?
            – AJGK
            Aug 19 at 5:39










          • Of course, it is.
            – Nosrati
            Aug 19 at 5:45















          Since I found the radius of convergence to be [-2, 0), would I then plug in (-2) such as f(x) = -ln(-(-2)) = -ln(2) ?
          – AJGK
          Aug 19 at 5:39




          Since I found the radius of convergence to be [-2, 0), would I then plug in (-2) such as f(x) = -ln(-(-2)) = -ln(2) ?
          – AJGK
          Aug 19 at 5:39












          Of course, it is.
          – Nosrati
          Aug 19 at 5:45




          Of course, it is.
          – Nosrati
          Aug 19 at 5:45












           

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