Tuesday, March 07, 2017

Harmonic series without number 9 is convergent

There are tons of proofs that harmonic series diverges. What if we have a harmonic series where we get rid of all the numbers which contains $9$, will it still converge. For single digit there are $8$ numbers which do not contain number $9$. Single digit numbers $8$ Two digit numbers $8 \times 9 $ Three digit numbers $8 \times 9 \times 9 = 8\cdot 9^2$ 4 digit numbers $latex 8\cdot 9^3$ ... Also notice that largest fraction in harmonic series Two digits $\frac{1}{10}$ Three digits $\frac{1}{100}$ Four digits $\frac{1}{1000}$ So if we replace the terms of harmonic with these fractions we will get an upper bound on the harmonic series devoid of number $9$ It is $\frac{1}{1}\cdot 8+\frac{1}{10}\cdot 8 \cdot 9+\frac{1}{100}\cdot 8 \cdot 9^2+\cdots+..$ and $\Rightarrow 8\cdot \frac{1}{1}+8\cdot \frac{9}{10}+8\cdot \frac{9^2}{10^2}+8\cdot \frac{9^3}{10^3}+\cdots$ This is a geometric series and the sum is $\frac{8}{1-\frac{9}{10}} = 80$. Thus the geometric series with common ratio of $\frac{9}{10}$ is convergent.

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