How can we modify the harmonic series to make it convergent?

Another way to modify the harmonic series

Upper left: Plot from 1 - 9, Upper right: Plot from 10 - 99, Lower left: Plot from 100 - 999, Lower Right: Plot from 1000 - 9999.

Just exactly how many terms are removed though?

The proof

Extra tidbits

The harmonic series is considered a classic textbook example of a series that seems to converge but is actually divergent. For those that need a reminder, the harmonic series is: \[\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots.\]Now here's an interesting question:Disregarding trivial answers such as "remove all the terms", this question is actually pretty hard to answer. Say we removed every second term starting from the first (i.e. 1, 1/3, 1/5, etc). Would that work?\[\sum_{\substack{n=1 \\ n\ is\ even}}^{\infty}\frac{1}{n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\cdots=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots\right).\]From above, it seems like it would not work since it just produces half of the original series (which is not a very well defined quantity since the original series was divergent anyway).Instead of removing every second term, let us(i.e. remove 1/9, 1/19, 1/29, etc). This series is called the Kempner series: \[1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{8}+\frac{1}{10}+\cdots.\]Although removing every term with a 9 doesn't seem like a lot, this bizarrely modified harmonic series is actually convergent (it is also convergent if you choose to remove any other digit/string of digits such as "0", "42", or "8675309")!We can now state the formal question as:Consider the series: \[a_{n}=\begin{cases}1/n & if\ n\ does\ not\ contain\ the\ digit\ 9 \\ 0 & otherwise\end{cases}\] Is $\sum a_n$ convergent?Instead of starting the proof right away, let's look at why this modification to the harmonic series might make it convergent.The plots below read 1 whenever an integer contains the digit 9, and 0 otherwise.As you can see, the plots have a nice "fractal" quality to them, in that each plot looks like it makes up parts of the next plot. Although the actual number of terms removed by the modification seems fewer than doing something like removing every second term, we can see that there are always points in the series where a long stretch of consecutive terms are removed (in fact, name a number and we can always find a point in the series where the number of consecutive terms removed is greater than that number). As the number of digits in the terms get large, the number of terms we remove increase drastically. As an example, most integers of say 100 digits will have a 9 in them, leading to about 99.99997% of the 100 digit terms being deleted. It is the quirky effect of removing consecutive terms as well as the fractal quality of the series that will allow us to see and prove its convergence.To do this systematically, we'll exploit the pattern we saw in the plots above. That is, since the plots seem to exhibit fractal qualities in the ranges ([1,9], [10,99], [100, 999] ...) that they are plotted, we'll look at the number of terms removed inside these ranges.For the interval [1,9], only. Note that this range starts at 1 because the harmonic series starts at n=1.For the interval [10, 99], we have 1 term removed from each of 10 - 19, 20 - 29, 30 - 39, ... , 80 - 89. Then we have 10 terms removed from 90 - 99. In total, we have 8 terms removed from 10 - 89, and a further 10 terms removed from 90 - 99. This makes for 8*1 + 10 =For the interval [100, 999], there is one term removed from 100 - 109, 18 terms removed from 110 - 199, 1 term removed from 200 - 219, 18 terms removed from 220 - 299, and so on, with the last being 18 terms removed from 800 - 899 terms. Finally, we have 100 terms removed from [900, 999]. In total this is 8*1 + 8*18 + 10² =Perhaps you are seeing a pattern here. Let's try to make out what that pattern is. Instead of looking at the number of terms removed, let's look at theFor the interval [1,9], we keep 9 - 1 =For the interval [10, 99], we keep 90 - 18 =For the interval [100, 999], we keep 900 - 252 =The pattern is 8, 72, 648. Dividing by 8, we have the pattern 1, 9, 81. We can see that the next number in the sequence will be 81*9 = 729, making the number of terms kept 8*729 = 5832. So the number of terms kept in a given interval $[10^n,10^{n+1}-1]$ is $8\cdot 9^n$. We will assume this knowledge (though it can be proven by standard methods of mathematical induction) for the proof below.Before we start the proof, we now can see why our modification to the series is so effective. Consider all integers containing 100 digits. There are $9\cdot 10^{99}$ of them. The number of terms that are kept is $8\cdot 9^{99}$. The fraction of terms that are kept is thus \[\frac{8\cdot 9^{99}}{9\cdot 10^{99}}\approx0.0000262335.\] We took out a lot of terms!Consider the series: \[a_{n}=\begin{cases}1/n & if\ n\ does\ not\ contain\ the\ digit\ 9 \\ 0 & otherwise\end{cases}\] Is $\sum a_n$ convergent?We know from the section above that the number of terms that are not 0 in $a_n$ on an interval $[10^n,10^{n+1}-1]$ is $8\cdot 9^n$. Moreover, the largest term of the harmonic series in each interval $[10^n,10^{n+1}-1]$ is $\frac{1}{10^n}$.Therefore, we can provide an upper bound for $\sum a_{n}$:\[\sum a_{n}here (under teaching and Science One).2. R. Baillie,American Mathematical Monthly, 372-374.3. T. Schmelzer and R. Baillie,American Mathematical Monthly, 525-540.