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

Answered by MJS last updated on 19/Mar/20

$$x=\frac{1}{3}+\frac{8}{9}+1+\frac{64}{81}+\frac{125}{243}+...=S_{\mathrm{\infty }}$$$$S_n=\underset{k=1}{\sum ^n}\frac{k^3}{3^k}=\frac{33}{8}-(\frac{n^3}{2}+\frac{9n^2}{4}-\frac{9n}{2}-\frac{33}{8})\times \frac{1}{3^n}$$$$\Rightarrow S_{\mathrm{\infty }}=\frac{33}{8}$$

Commented by naka3546 last updated on 19/Mar/20

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