# function to which power series converges within radius of convergence

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?

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

## 2 Answers

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 $|x|<1$.

• 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? – user486635 Apr 25 '18 at 3:09
• Yes. You can derive that series term by term when you are inside of the radius of convergence and convergence is uniform. – Pablo Herrera Apr 25 '18 at 3:11
• Thank you so much -- I understand now! – user486635 Apr 25 '18 at 3:14