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Question Number 127889 by A8;15: last updated on 02/Jan/21

Answered by mindispower last updated on 02/Jan/21

=Σ_(n≥1) (1/(5^n n)) (1/(1−x))=Σ_(n≥0) x^n ⇒∫_0 ^t (1/(1−x))dx=^� Σ_(n≥1) (t^n /n),,t=(1/5)⇒ ∫_0 ^(1/5) (dx/(1−x))=Σ_(n≥1) (1/(n5^n ))=−ln(1−(1/5))=ln((5/4))

$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{5}^{{n}} {n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}}=\underset{{n}\geqslant\mathrm{0}} {\sum}{x}^{{n}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{1}}{\mathrm{1}−{x}}{dx}\hat {=}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{t}^{{n}} }{{n}},,{t}=\frac{\mathrm{1}}{\mathrm{5}}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \frac{{dx}}{\mathrm{1}−{x}}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}\mathrm{5}^{{n}} }=−{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)={ln}\left(\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$

Commented by A8;15: last updated on 03/Jan/21

Thanks sir. Happy New Year

$$\mathrm{Thanks}\:\mathrm{sir}.\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Year}} \\ $$

Commented by mindispower last updated on 03/Jan/21

withe pleasur happy new year

$${withe}\:{pleasur} \\ $$$${happy}\:{new}\:{year} \\ $$

Answered by mr W last updated on 02/Jan/21

ln (1+x)=x−(x^2 /2)+(x^3 /3)−(x^4 /4)+... ln (1−x)=−x−(x^2 /2)−(x^3 /3)−(x^4 /4)−... ⇒ln (1/(1−x))=x+(x^2 /2)+(x^3 /3)+(x^4 /4)+... with x=(1/5) ⇒(1/5)+(1/(5^2 ×2))+(1/(5^3 ×3))+(1/(5^4 ×4))+...=ln (5/4)

$$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+... \\ $$$$\mathrm{ln}\:\left(\mathrm{1}−{x}\right)=−{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}−... \\ $$$$\Rightarrow\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}={x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+... \\ $$$${with}\:{x}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} ×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} ×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{4}} ×\mathrm{4}}+...=\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$

Commented by A8;15: last updated on 03/Jan/21

Thanks Mr. W. Happy New Year

$$\mathrm{Thanks}\:\mathrm{Mr}.\:\mathrm{W}.\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Year}} \\ $$

Commented by mr W last updated on 03/Jan/21

happy new year too!

$${happy}\:{new}\:{year}\:{too}! \\ $$

Answered by Dwaipayan Shikari last updated on 03/Jan/21

(1/5)+(1/(5^2 .2))+(1/(5^3 .3))+... =−log(1−(1/5))=log((5/4)) Generally Σ_(n=1) ^∞ (1/(nk^n ))=log((k/(k−1)))

$$\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} .\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} .\mathrm{3}}+... \\ $$$$=−{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)={log}\left(\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$${Generally}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{nk}^{{n}} }={log}\left(\frac{{k}}{{k}−\mathrm{1}}\right) \\ $$

Commented by A8;15: last updated on 03/Jan/21

Thanks sir. Happy New Year

$$\mathrm{Thanks}\:\mathrm{sir}.\:\boldsymbol{\mathrm{Happy}}\:\boldsymbol{\mathrm{New}}\:\boldsymbol{\mathrm{Year}} \\ $$

Commented by Dwaipayan Shikari last updated on 03/Jan/21

Have a great year!

$${Have}\:{a}\:{great}\:{year}! \\ $$

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