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Question Number 151247 by mathdanisur last updated on 19/Aug/21

(1/5) + (1/(5^2 ∙2)) + (1/(5^3 ∙3)) + (1/(5^4 ∙4)) + ... = ?

$$\frac{\mathrm{1}}{\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} \centerdot\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} \centerdot\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{4}} \centerdot\mathrm{4}}\:+\:...\:=\:? \\ $$

Answered by qaz last updated on 19/Aug/21

Σ_(n=1) ^∞ (1/(n5^n ))=−ln(1−(1/5))=ln(5/4)

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n5}^{\mathrm{n}} }=−\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)=\mathrm{ln}\frac{\mathrm{5}}{\mathrm{4}} \\ $$

Commented by mathdanisur last updated on 19/Aug/21

Thankyou Ser, but ln(5/4) or ln(4/5).?

$$\mathrm{Thankyou}\:\mathrm{Ser},\:\mathrm{but}\:\mathrm{ln}\frac{\mathrm{5}}{\mathrm{4}}\:\mathrm{or}\:\mathrm{ln}\frac{\mathrm{4}}{\mathrm{5}}.? \\ $$

Commented by naka3546 last updated on 19/Aug/21

ln (5/4)

$$\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$

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