Finding a sum involving roots of a quadratic equation

Question

If $\alpha,\beta$ are roots of the equation $x^2-2x-7=0$ and $$S_r=\left(\frac{r}{\alpha ^r}+\frac{r}{\beta ^r}\right)$$ then find the value of $$\lim _{n \to \infty} \sum _{r=1} ^n S_r$$ I am unable to telescope $S_r$, though it looks possible. Thanks.

Answer 1

Since $\alpha,\beta=1\pm 2\sqrt 2,\alpha\beta=-7$, $$S_r=r\cdot\frac{\alpha^r+\beta^r}{(\alpha\beta)^r}=r\left(\frac{1-2\sqrt 2}{-7}\right)^r+r\left(\frac{1+2\sqrt 2}{-7}\right)^r$$ So, $$\lim_{n\to\infty}\sum_{r=1}^nS_r=\lim_{n\to\infty}\left(\sum_{r=1}^{n}r\left(\frac{1-2\sqrt 2}{-7}\right)^r+\sum_{r=1}^{n}r\left(\frac{1+2\sqrt 2}{-7}\right)^r\right)$$ Here, use $$\lim_{n\to\infty}\sum_{r=1}^{n}ra^r=\frac{a}{(1-a)^2}$$ for $|a|\lt 1$.

Answer 2

$\alpha=1-2\sqrt 2;\;\beta=1-2\sqrt 2$

$S_r=\dfrac{r}{\left(1-2\sqrt 2\right)^r} + \dfrac{r}{\left(1+2\sqrt 2\right)^r}$

Now remember that

$s_n=\sum\limits_{r = 1}^n r{a^r} = \dfrac{a}{{{{\left( {a - 1} \right)}^2}}} + \dfrac{a^{n + 1}\left( {an - n - 1} \right)}{{\left( a - 1 \right)}^2}$

both $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ are less then $1$, so the limit of $s_n$ as $n\to \infty $ is $\dfrac{a}{\left( a - 1 \right)^2}$ and the series value is $\dfrac{\dfrac{1}{\alpha}}{\left( \dfrac{1}{\alpha} - 1 \right)^2}+\dfrac{\dfrac{1}{\beta}}{\left( \dfrac{1}{\beta} - 1 \right)^2}=\dfrac{1}{4}$