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.
Labels: 9, Convergence, Harmonic series
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