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

Find: (1/4) + (1/8) + (3/(64)) + (4/(256)) + ... + = ?

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

Commented by Frix last updated on 05/Jun/24

=(1/4)+(2/(16))+(3/(64))+(4/(256))+...= =Σ_(k=1) ^∞ (k/4^k ) =lim_(n→∞) ((4^(n+1) −(3n+4))/(9×4^n )) =(4/9) f(n)=Σ_(k=1) ^n (k/a^k ) =((a^(n+1) −(n+1)a+n)/((a−1)^2 a^n )) ∀a>1: lim_(n→∞) f(n) =(a/((a−1)^2 ))

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

Commented by hardmath last updated on 06/Jun/24

thank you dear professor

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

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