I am asked to find the power series expansion for $f(x)$ on the interval(-1,1) $$f(x)=\frac{-(x+1)}{(x-1)^3}$$

I get $1+4x+9x^2+16x^3$


$$ \sum_{k=2}^\infty\ k^2x^{k-1}$$

My question is how to write the summation for $$ \sum_{k=0}^\infty\ $$ I'm not sure how to write the summation to include the first term since the first term has no $x$.

Also I am asked to evaluate the power series at $1/2$ in $(-1,1)$ Does that mean just set the c value which was $0$ in the Maclaurin series and find the Taylor series for $1/2$?

Any help would be greatly appreciated.

Thank you,


1 Answer 1


$$\sum_{k=2}^{\infty}k^2x^{k-1}\equiv \sum_{k=0}^\infty (k+1)^2x^{k}$$

I think you are correct about the second thing. Perhaps they mean power series about $x=0.5$ which then means you apply the Taylor Formula about $x=0.5.$

  • 1
    $\begingroup$ The second sum should start at $k=1$, not $k=0$ $\endgroup$
    – saulspatz
    Mar 24, 2018 at 23:52
  • 1
    $\begingroup$ I'm pretty sure that Harry Alli got it right since when $k=0$ we have $(0+1)^{2}x^0=1$ and $k=1$ we have $(1+1)^2x^1=4x$, so why would the summation start at $k=1$ not $k=0$?@saulspatz $\endgroup$
    – jack
    Mar 25, 2018 at 0:06

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