# Convergence of a Series

Let $N=\{n_1, n_2, ...\}$ Be a sequence of all positive integers whose decimal representation does not contain the digit $0$. Hence 45 is in $N$ but $10$ is not.

Show that $\sum _{k=1}^{\infty } \frac{1}{n_k}$ converges. Also show that the sum is less than $90$.

I've tried some of the usual "tests for convergence" like the ratio test but they come out inconclusive or they don't apply. What I was trying to work with was the harmonic series
$\sum _{k=1}^{\infty } \frac{1}{n}$
and then subtract out all the numbers that contain the digit zero but it wasn't working very well...

There are $9^n$ numbers with exactly $n$ digits none of which is $0$, and each of these is greater than or equal to $10^{n-1}$, with equality holding only once. This bound is enough to show that the sum is less than $90$ using a geometric series.
• I feel like I may be thinking to far into this. Since $9^n10^{n-1} \leq \sum _{k=1}^{9^n} n_k$ then $\sum _{n=1}^{\infty } \frac{1}{\sum _{k=1}^{9^n} n_k} \leq \sum _{n=1}^{\infty } \frac{1}{9^n10^{n-1}}$ I'm not sure where to go from here – Susie Q Apr 12 '11 at 2:26
• @Susie Q: $$\sum_{n_k\text{ has }n\text{ digits}}\frac{1}{n_k}\lt 9^n\frac{1}{10^{n-1}}.$$ – Jonas Meyer Apr 12 '11 at 2:40
• I'm sorry, I don't understand how to relate this to a geometric series... Can I say $9^n \frac{1}{10^{n-1}} \leq 10^n \frac{1}{10^{n-1}} = 10$ ? – Susie Q Apr 12 '11 at 2:51
• @Susie Q: That inequality is true but unhelpful. Instead I recommend using $\displaystyle{9^n\frac{1}{10^{n-1}}=9\left(\frac{9}{10}\right)^{n-1}}$. – Jonas Meyer Apr 12 '11 at 2:56