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Find what the following number approaches: $$\frac{6}{5}\times\frac{26}{25}\times\frac{626}{625}\times\frac{390626}{390625}\times...$$

What I found is that each fraction can be simplified as:$$1+\frac{1}{5^n}$$ Where $n$ is $1,2,4,8$ and so on (instead of $1,2,3,4,\ldots$)

What should I do next?

Note: Please use junior high school math if possible

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  • $\begingroup$ @Damien Thanks for the notice. $\endgroup$ – Cyh1368 Jan 27 at 9:39
  • $\begingroup$ or do u mean the reciprocal of what you have written? $\endgroup$ – Aatmaj Jan 27 at 9:41
  • $\begingroup$ Hint: Notice that the product is in fact $$ \sum_{i=0}^\infty 5^{-i} $$ you can observe it by just expand the product. Take one term in each product and you will see that. $\endgroup$ – macton Jan 27 at 9:42
  • $\begingroup$ The same idea math.stackexchange.com/q/622321/399263 $\endgroup$ – zwim Jan 27 at 11:27
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You have

$$\left(1+\frac{1}{5^1}\right)\left(1+\frac{1}{5^2}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}\right)\left(1+\frac{1}{5^4}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=\left(1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}+\frac{1}{5^6}+\frac{1}{5^7}\right)\left(1+\frac{1}{5^8}\right)\cdots$$

$$=1+\frac{1}{5^1}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}+\frac{1}{5^6}+\frac{1}{5^7}+\frac{1}{5^8}+\cdots$$

$$= \frac{5}{5-1} \\= 1+\frac{1}{4}$$

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Put $x=\frac{1}{5}$ then we have $$\lim_{n \to \infty}(1+x)(1+x^2)(1+x^4)(1+x^8)..(1+x^{2^n})$$ now multiply $1-x$ in numerator and denominator and use $(1-x^{2^k})(1+x^{2^k})=1-x^{2^{k+1}}$ repeatedly and see the magic!

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