# Find what a Infinite product number approaches

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

• @Damien Thanks for the notice. – Cyh1368 Jan 27 at 9:39
• or do u mean the reciprocal of what you have written? – Aatmaj Jan 27 at 9:41
• 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. – macton Jan 27 at 9:42
• The same idea math.stackexchange.com/q/622321/399263 – zwim Jan 27 at 11:27

You have

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

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

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

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

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

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!