This post is motivated by this cartoon
which I found at a Evelyn Lamb’s post on an AMS Blog, this fun Forbes math post by Kevin Kundson and by a June 2015 article in Mathematics Magazine by R. John Ferdinands called Selective Sums of an Infinite Series.
Here is the following question: start with a divergent series of positive terms which form a decreasing (non-increasing) sequence which tends to zero, say, 
. Now how does one select a subset of series terms to delete so as to obtain a convergent series? The Kundson article shows that one can do this with the harmonic series by, say, deleting all numbers that contain a specific digit (say, 9). I’ll talk about the proof here. But I’d like to start more basic and to bring in language used in the Ferdinands article.
So, let’s set the stage: we will let 
denote the divergent sum in question. All terms will be positive, 
for all 
and 
. Now let 
represent a sequence where 
for all ; then 
is called a selective sum of . I’ll call the the selecting sequence and, from the start, rule out selecting sequences that are either eventually 1 (which means that the selected series diverges since the original series did) or eventually zero (just a finite sum).
Now we’ll state a really easy result:
There is some non-eventually constant such that converges. Here is why: because , for each 
one can find a maximal index 
so that 
. Now select 
if 
and 
otherwise. Then 
and therefore the selected series converges by comparison with a convergent geometric series.
Of course, this result is petty lame; this technique discards a lot of terms. A cheap way to discard “fewer” terms (“fewer” meaning: in terms of “set inclusion”): Do the previous construction, but instead of using 
use 
where 
is a positive integer of choice. Note that 
Here is an example of how this works: Consider the divergent series 
and the convergent geometric series 
Of course 
so 
but then for 
we have 
. So for 
. But 
because 
. The next non-zero selection coefficient is 
as 
.
Now playing with this example, we see that 
for 
but not for 
. So 
for 
and 
. So the first few 
are 
. Of course the gap between the grows as does.
Now let’s get back to the cartoon example. From this example, we’ll attempt to state a more general result.
Claim: given 
where if contains a 9 as one of its digits, then converges. Hint on how to prove this (without reading the solution): count the number of integers between 
and 
that lack a 9 as a digit. Then do a comparison test with a convergent geometric series, noting that every term 
is less than or equal to 
.
How to prove the claim: we can start by “counting” the number of integers between 0 and that contain no 9’s as a digit.
Between 0 and 9: clearly 0-8 inclusive, or 9 numbers.
Between 10 and 99: a moment’s thought shows that we have 
numbers with no 9 as a digit (hint: consider 10-19, 20-29…80-89) so this means that we have 
numbers between 0 and 99 with no 9 as a digit.
This leads to the conjecture: there are 
numbers between 0 and 
with no 9 as a digit and 
between 
and 
with no 9 as a digit.
This is verified by induction. This is true for 
Assume true for 
. Then to find the number of numbers without a 9 between 
and 
we get 
which then means we have 
numbers between 0 and 
with no 9 as a digit. So our conjecture is proved by induction.
Now note that 
This establishes that 
So it follows that 
and hence our selected sum is convergent.
Further questions: ok, what is going on is that we threw out enough terms of the harmonic series for the series to converge. Between terms and 
we allowed 
terms to survive.
This suggests that if we permit up to 
terms between and 
to survive (
fixed and positive) then we will have a convergent series. I’d be interested in seeing if there is an generalization of this.
But I am tried, I have a research article to review and I need to start class preparation for the upcoming spring semester. So I’ll stop here. For now. 🙂
