Getting $f(x) =sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}$, $|x|<1$. [duplicate]












1












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This question already has an answer here:




  • What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?

    2 answers




In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.



What is $f(x)$ itself?



Here is the Taylor series at $x=0$ for the function:



$$
f(x)=sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}
$$



I'm not sure about the $1$ in $n-1$.






calculus taylor-expansion







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marked as duplicate by Martin R, mrtaurho, Community♦ Jan 1 at 20:19


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    $begingroup$
    Use partial fractions to simplify the sum.
    $endgroup$
    – D.B.
    Jan 1 at 19:31










  • $begingroup$
    Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
    $endgroup$
    – LoveTooNap29
    Jan 1 at 19:31










  • $begingroup$
    Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
    $endgroup$
    – Martin R
    Jan 1 at 19:47










  • $begingroup$
    yes , thanks a lot
    $endgroup$
    – mahsa.e.1378
    Jan 1 at 20:09
















1












$begingroup$



This question already has an answer here:




  • What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?

    2 answers




In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.



What is $f(x)$ itself?



Here is the Taylor series at $x=0$ for the function:



$$
f(x)=sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}
$$



I'm not sure about the $1$ in $n-1$.






calculus taylor-expansion







share|cite|improve this question











$endgroup$



marked as duplicate by Martin R, mrtaurho, Community♦ Jan 1 at 20:19


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    Use partial fractions to simplify the sum.
    $endgroup$
    – D.B.
    Jan 1 at 19:31










  • $begingroup$
    Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
    $endgroup$
    – LoveTooNap29
    Jan 1 at 19:31










  • $begingroup$
    Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
    $endgroup$
    – Martin R
    Jan 1 at 19:47










  • $begingroup$
    yes , thanks a lot
    $endgroup$
    – mahsa.e.1378
    Jan 1 at 20:09














1












1








1





$begingroup$



This question already has an answer here:




  • What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?

    2 answers




In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.



What is $f(x)$ itself?



Here is the Taylor series at $x=0$ for the function:



$$
f(x)=sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}
$$



I'm not sure about the $1$ in $n-1$.






calculus taylor-expansion







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?

    2 answers




In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.



What is $f(x)$ itself?



Here is the Taylor series at $x=0$ for the function:



$$
f(x)=sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}
$$



I'm not sure about the $1$ in $n-1$.





This question already has an answer here:




  • What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?

    2 answers







calculus taylor-expansion




calculus taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 21:07









mrtaurho

4,05921234




4,05921234










asked Jan 1 at 19:28









mahsa.e.1378mahsa.e.1378

71




71




marked as duplicate by Martin R, mrtaurho, Community♦ Jan 1 at 20:19


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Martin R, mrtaurho, Community♦ Jan 1 at 20:19


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    Use partial fractions to simplify the sum.
    $endgroup$
    – D.B.
    Jan 1 at 19:31










  • $begingroup$
    Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
    $endgroup$
    – LoveTooNap29
    Jan 1 at 19:31










  • $begingroup$
    Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
    $endgroup$
    – Martin R
    Jan 1 at 19:47










  • $begingroup$
    yes , thanks a lot
    $endgroup$
    – mahsa.e.1378
    Jan 1 at 20:09














  • 1




    $begingroup$
    Use partial fractions to simplify the sum.
    $endgroup$
    – D.B.
    Jan 1 at 19:31










  • $begingroup$
    Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
    $endgroup$
    – LoveTooNap29
    Jan 1 at 19:31










  • $begingroup$
    Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
    $endgroup$
    – Martin R
    Jan 1 at 19:47










  • $begingroup$
    yes , thanks a lot
    $endgroup$
    – mahsa.e.1378
    Jan 1 at 20:09








1




1




$begingroup$
Use partial fractions to simplify the sum.
$endgroup$
– D.B.
Jan 1 at 19:31




$begingroup$
Use partial fractions to simplify the sum.
$endgroup$
– D.B.
Jan 1 at 19:31












$begingroup$
Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
$endgroup$
– LoveTooNap29
Jan 1 at 19:31




$begingroup$
Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
$endgroup$
– LoveTooNap29
Jan 1 at 19:31












$begingroup$
Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
$endgroup$
– Martin R
Jan 1 at 19:47




$begingroup$
Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
$endgroup$
– Martin R
Jan 1 at 19:47












$begingroup$
yes , thanks a lot
$endgroup$
– mahsa.e.1378
Jan 1 at 20:09




$begingroup$
yes , thanks a lot
$endgroup$
– mahsa.e.1378
Jan 1 at 20:09










1 Answer
1






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3












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Hint. One may recall that
$$
sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
$$ Then writing
$$
frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
$$ may help.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Hint. One may recall that
    $$
    sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
    $$     Then writing
    $$
    frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
    $$     may help.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Hint. One may recall that
      $$
      sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
      $$       Then writing
      $$
      frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
      $$       may help.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Hint. One may recall that
        $$
        sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
        $$         Then writing
        $$
        frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
        $$         may help.






        share|cite|improve this answer









        $endgroup$



        Hint. One may recall that
        $$
        sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
        $$         Then writing
        $$
        frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
        $$         may help.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 19:33









        Olivier OloaOlivier Oloa

        108k17176293




        108k17176293















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