Sum until Infinity

Calculus Level 3

If given that $\displaystyle \sum_{1}^{\infty} \frac{x^{2}}{16^{x}} = \frac{a}{b}$ , find the value of $\sqrt{a^{2}+b^{2}}.$

Note: $\frac{a}{b}$ is in the simplest ratio with no common multiples. And round off the final answer up to 3 decimal places.

1 solution

Reineir Duran
Jan 21, 2016

My solution is quite easy to understand. Let

$\displaystyle S = \frac{1^2}{16} + \frac{2^2}{16^2} + \frac{3^2}{16^3} + \frac{4^2}{16^4} + ...$ $(1)$

$\displaystyle \Longrightarrow 16S = 1^2 + \frac{2^2}{16} + \frac{3^2}{16^2} + \frac{4^2}{16^3} + ...$ $(2)$

Now, $(2) - (1)$ gives

$\displaystyle 15S = 1 + \frac{3}{16} + \frac{5}{16^2} + \frac{7}{16^2} + ...$ $(3)$

$\displaystyle \Longrightarrow 16(15S) = 16 + 3 + \frac{5}{16} + \frac{7}{16^2} + ...$ $(4)$

Moreover, $(4) - (3)$ gives

$\displaystyle \Longrightarrow 225S = 18 + \frac{2}{16} + \frac{2}{16^2} + ... \Longleftrightarrow 225S = 18 + \frac{2}{15} = \frac{272}{15}$

Thus, $\displaystyle S = \frac{272}{3375}$ and $\displaystyle \sqrt{a^2 + b^2} = \sqrt{272^2 + 3375^2} \approx 3385.943$ .