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. 🙂