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Question Number 75828 by 21042004 last updated on 18/Dec/19

Σ_(n=1) ^∞ (1/(10^n ))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{10}^{{n}} } \\ $$

Commented by mathmax by abdo last updated on 18/Dec/19

Σ_(n=1) ^∞  (1/(10^n )) =Σ_(n=0) ^∞ ((1/(10)))^n −1 =(1/(1−(1/(10))))−1 =((10)/9)−1 =(1/9)

$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{10}^{{n}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{10}}\right)^{{n}} −\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}}−\mathrm{1}\:=\frac{\mathrm{10}}{\mathrm{9}}−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{9}} \\ $$

Answered by MJS last updated on 18/Dec/19

=.1+.01+.001+...=.1^(.) =(1/9)

$$=.\mathrm{1}+.\mathrm{01}+.\mathrm{001}+...=.\overset{.} {\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$

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