The question is to determine the radius of convergence of as well as the function to which the following series converges to (within that radius): $\sum_{j=1}^{\infty} (j + 1)(j + 2)x^j$.

I was able to determine that it converges whenever $|x| < 1$, but I wasn't able to determine to what function it will converge (or think of a starting point for that). Any tips?

  • 2
    $\begingroup$ Hint: this series is the second derivative of a much nicer series. $\endgroup$ – Alex S Apr 25 '18 at 3:02

Doing integration may help: \begin{align*} \int_{0}^{x}\sum_{j=1}(j+1)(j+2)t^{j}dt&=\sum_{j=1}(j+2)t^{j+1}\bigg|_{t=0}^{t=x}\\ &=\sum_{j=1}(j+2)x^{j+1}, \end{align*} and once more \begin{align*} \int_{0}^{x}\sum_{j=1}(j+2)t^{j+1}dt=\sum_{j=1}x^{j+2}=\dfrac{x^{3}}{1-x}, \end{align*} so the series is the twice derivative of $x^{3}/(1-x)$.


Think about derivatives of geometric series: $$\sum_{j=1}^{\infty}x^{j+2}=\frac{x^3}{1-x}$$

around $|x|<1$.

  • $\begingroup$ does that mean automatically that the original series converges to $\frac{x^3}{1-x}''$? surely that isn't always the case (though your hint is extremely helpful)? does the fact that the original series converges uniformly to some function in its radius of convergence help? $\endgroup$ – user486635 Apr 25 '18 at 3:09
  • 1
    $\begingroup$ Yes. You can derive that series term by term when you are inside of the radius of convergence and convergence is uniform. $\endgroup$ – Pablo Herrera Apr 25 '18 at 3:11
  • $\begingroup$ Thank you so much -- I understand now! $\endgroup$ – user486635 Apr 25 '18 at 3:14

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