The Kempner series is a modification of the harmonic series where we remove all numbers that contain the digit $9$. Interestingly, this series converges.
I'm wondering whether it will still converge if instead of the harmonic series, we consider $\displaystyle \frac{1}{n^p}$.
We can work out some cases. For example, when $p > 1$, we can use a comparison test with the usual $\displaystyle \frac{1}{n^p}$ with no terms removed, and so it converges. On the other hand, if we let $p = 0$ or $p < -1$, then since there are infinitely many terms, so it clearly diverges.
What about when $-1<p<1?$
My intuition is that it will diverge, but I'm not so certain. How do we go about proving this?
Your modified Kempner series converges iff $p>\frac{\ln 9}{\ln 10}$.
For $n\geq 0$ let $A_n$ be the set of integers between $10^n$ and $10^{n+1}-1$ (inclusive) that don't have any $9$'s in their decimal expansion.
Note that $\sum_{n=0}^\infty\frac{1}{10^{(n+1)p}}|A_n|\leq \sum_{n=0}^\infty \sum_{k\in A_n}\frac{1}{k^p}\leq \sum_{n=0}^\infty\frac{1}{10^{np}}|A_n|$
But $|A_n| = 8\cdot9^n$, thus $\frac{8}{10^p}\sum_{n=0}^\infty\left(\frac{9}{10^{p}}\right)^n\leq \sum_{n=0}^\infty \sum_{k\in A_n}\frac{1}{k^p}\leq \sum_{n=0}^\infty\left(\frac{9}{10^{p}}\right)^n$
Therefore, the series converges iff $\frac{9}{10^{p}}<1$, that is $p>\frac{\ln 9}{\ln 10}$.