<s>[INST] Does removing all numbers in the harmonic series with a units digit of 9 affect the series? The harmonic sequence is $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\ldots$ diverges. There is a simple reason why:
$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\ldots$ is obviously greater than $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8}+\frac{1}{8}+\frac{1}{8}\ldots$ , which is infinte.
But, if you remove all numbers that include a "9", the series converges. The reason for this is that as the amount of digits in the number increases, it will be harder and harder to avoid the digit "9", and it will slowly become impossibly hard.
I was wondering that if you remove every number with a units digit of 9, will the series converge or diverge? 
 [/INST] The series $\sum\limits_{n=1}^\infty\frac{1}{2n}$ diverges because $\sum\limits_{n=1}^N\frac{1}{2n}=\frac12\sum\limits_{n=1}^N\frac{1}{n}$.  You get rid of all odd units digits this way, still diverging.
The series $\sum\limits_{n=1}^\infty\dfrac{1}{10^k n}$ for any fixed $k\in\mathbb N$ gives you $0$s in the first $k$ places of each number, and still diverges.
You could also take away all terms with leading digit $9$ and still get a divergent series.  In fact, only taking terms with leading digit $1$ yields 
$$1+\frac1{10}+\cdots+\frac{1}{19}+\frac{1}{100}+\cdots+\frac{1}{199}+\cdots\geq 1+10\cdot\frac{1}{20}+100\cdot\frac1{200}+\cdots.$$
Combining this with the previous trick of multiplying by $10^k$, you could have only leading digits $1$, with an arbitrary but fixed number of ending $0$s, and still diverge.
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