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Question Number 208105 by hardmath last updated on 05/Jun/24

$$\mathrm{Find}:$$$$\frac{1}{4}+\frac{1}{8}+\frac{3}{64}+\frac{4}{256}+...+=?$$

Commented by Frix last updated on 05/Jun/24

$$=\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+...=$$$$=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{k}{4^k}=\underset{n\to \mathrm{\infty }}{\lim }\frac{4^{n+1}-(3n+4)}{9\times 4^n}=\frac{4}{9}$$$$$$$$f\left(n\right)=\underset{k=1}{\sum ^n}\frac{k}{a^k}=\frac{a^{n+1}-(n+1)a+n}{{(a-1)}^2{a}^n}$$$$\mathrm{\forall }a>1:\underset{n\to \mathrm{\infty }}{\lim }f\left(n\right)=\frac{a}{{(a-1)}^2}$$

Commented by hardmath last updated on 06/Jun/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

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