Question-85105

Question Number 85105 by naka3546 last updated on 19/Mar/20

Answered by MJS last updated on 19/Mar/20

x = \frac{1}{3} + \frac{8}{9} + 1 + \frac{64}{81} + \frac{125}{243} + \dots = S_{\infty}

S_{n} = \underset{k = 1}{\sum^{n}} \frac{k^{3}}{3^{k}} = \frac{33}{8} - (\frac{n^{3}}{2} + \frac{9 n^{2}}{4} - \frac{9 n}{2} - \frac{33}{8}) \times \frac{1}{3^{n}}

\Rightarrow S_{\infty} = \frac{33}{8}

Commented by naka3546 last updated on 19/Mar/20