My power series is:
$$\sum \limits_{n=2}^{\infty} (\frac{1}{n(n-1)})x^n$$
Now I calculate the radius of convergence using the quotient criterion:
$$\lim\limits_{n\to\infty}|\frac{\frac{1}{(n+1)(n+1-1)}x^{n+1}}{\frac{1}{n(n-1)}x^n}|\rightarrow |x|\lim\limits_{n\to\infty}\frac{n-1}{n+1}\rightarrow |x|\cdot 1\rightarrow |x|<1\rightarrow radius\text{ } of\text{ } convergence\text{ r }=1$$
How can I use this to calculate the value of this series $$\sum \limits_{n=2}^{\infty} \frac{1}{n(n-1)3^n}$$ ?
I'm grateful for any help!
Best Answer
HINT
Consider the following power series (known as geometric series) which converges whenever $|x| < 1$: \begin{align*} \sum_{n=0}^{\infty}x^{n} = 1 + x + x^{2} + x^{3} + \ldots = \frac{1}{1 - x} \end{align*}
Then integrate it twice: \begin{align*} & \int_{0}^{x}\left(\sum_{n=0}^{\infty}y^{n}\right)\mathrm{d}y = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \ldots = \sum_{n=1}^{\infty}\frac{x^{n}}{n}\\\\ & \int_{0}^{x}\left(\sum_{n=1}^{\infty}\frac{y^{n}}{n}\right)\mathrm{d}y = \frac{x^{2}}{2} + \frac{x^{3}}{2\times 3} + \frac{x^{4}}{3\times 4} + \frac{x^{5}}{4\times 5} + \ldots = \sum_{n=2}^{\infty}\frac{x^{n}}{n(n-1)} \end{align*}
Based on the provided suggestion, can you take it from here?