Prove that this subserie of the harmonic series is convergent. [duplicate]

Solution 1:

lets consider another sequence that bounds this sequence from above let $a_i$ be a sequence with $a_i = 10^{|n_i|-1}$, where $|n_i|$ is the number of digits of $n_i$, obviosuly $a_i \leq n_i \implies \frac{1}{a_i} \geq \frac{1}{n_i} $ (with equality holding if and only if $n_i = 10^{d-1}$) and so $\sum \frac{1}{n_i} \leq \sum \frac{1}{a_i}$

now lets count all $a_i = 10^{d-1}$ for each d, each $a_i$ corresponds to an $n_i$ which is any number with $d$ digits that doesn't contain a zero, so their counts are equal and are equal to $9^d$, so the sum of all $\frac{1}{a_i}$ where $a_i = 10^{d-1}$ is $\frac{9^d}{10^{d-1}} = 9*(\frac{9}{10})^{d-1}$ putting it all together for all number of digits $\sum_{d=1}^{\infty} = 9 \sum_{d=1}^{\infty}(\frac{9}{10})^{d-1}$ which converges as $9 \frac{1}{1 - \frac{9}{10}} = 90$

in summary $\sum \frac{1}{n_i} \leq \frac{1}{a_i} = 90$