Question-211919

Question Number 211919 by Spillover last updated on 24/Sep/24

Answered by Frix last updated on 24/Sep/24

x^(1+(3/2)+(4/4)+(5/8)+(6/(16))...) =x^5 1+Σ_(k=1) ^∞ ((k+2)/2^k ) =1+2Σ_(k=1) ^∞ (1/2^k ) +Σ_(k=1) ^∞ (k/2^k ) 2Σ_(k=1) ^∞ (1/2^k ) =2×1 Σ_(k=1) ^∞ (k/2^k ) =lim_(k→∞) (2−((k+2)/2^k )) =2

$${x}^{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{6}}{\mathrm{16}}…} ={x}^{\mathrm{5}} \\ $$$$\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}+\mathrm{2}}{\mathrm{2}^{{k}} }\:=\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}}{\mathrm{2}^{{k}} }\: \\ $$$$\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:=\mathrm{2}×\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}}{\mathrm{2}^{{k}} }\:=\underset{{k}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}−\frac{{k}+\mathrm{2}}{\mathrm{2}^{{k}} }\right)\:=\mathrm{2} \\ $$

Commented by Spillover last updated on 24/Sep/24

$${i}\:{got}\:\mathrm{1} \\ $$

Commented by Frix last updated on 24/Sep/24

I misread x(√(x^2 (√(x^3 (√(x^4 ...)))))) x((x((x((x...))^(1/4) ))^(1/3) ))^(1/2) =x^(1+(1/2)+(1/6)+(1/(24))+...) =x^(e−1) ∫_0 ^1 x^(e−1) dx=[(x^e /e)]_0 ^1 =(1/e)

$$\mathrm{I}\:\mathrm{misread}\:{x}\sqrt{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{4}} …}}} \\ $$$${x}\sqrt[{\mathrm{2}}]{{x}\sqrt[{\mathrm{3}}]{{x}\sqrt[{\mathrm{4}}]{{x}…}}}={x}^{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{24}}+…} ={x}^{\mathrm{e}−\mathrm{1}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\mathrm{e}−\mathrm{1}} {dx}=\left[\frac{{x}^{\mathrm{e}} }{\mathrm{e}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{e}} \\ $$

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