College Math Teaching

January 14, 2016

Trimming a divergent series into a convergent one

Filed under: calculus, induction, sequences, series — Tags: , , — collegemathteaching @ 10:28 pm

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, \sum^{\infty}_{k =1} \frac{1}{k} . 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 \sum a_k denote the divergent sum in question. All terms will be positive, a_{k} \geq a_{k+1} for all k and lim_{k \rightarrow \infty} a_k = 0 . Now let c_k represent a sequence where c_k \in \{0,1\} for all k ; then \sum c_ka_k is called a selective sum of \sum a_k . I’ll call the c_k 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 c_k such that \sum c_ka_k converges. Here is why: because lim_{k \rightarrow \infty} a_k = 0 , for each n \in \{1,2,3...\} one can find a maximal index n_j, n_j \notin \{n_1, n_2, ...n_{j-1} \} so that \frac{1}{2^n} > a_{n_j} . Now select c_k = 1 if k \in \{n_1, n_2, n_3,... \} and c_k =0 otherwise. Then \sum \frac{1}{2^k} > \sum c_ka_k 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 \frac{1}{2} use \frac{M}{M+1} where M is a positive integer of choice. Note that \sum^{\infty}_{k=1} (\frac{M}{M+1})^k = M

Here is an example of how this works: Consider the divergent series \sum \frac{1}{\sqrt{k}} and the convergent geometric series \sum (\frac{1000}{1001})^k Of course \frac{1000}{1001} < 1 so c_1 = 0 but then for k \in \{2,3,....4169 \} we have (\frac{1000}{1001})^k > \frac{1}{\sqrt{k}} . So c_k = 1 for k \in \{2,3,4,....4169 \} . But c_{4170} = 0 because (\frac{1000}{1001})^{4170} < \frac{1}{\sqrt{4170}}. The next non-zero selection coefficient is c_{4171} as (\frac{1000}{1001})^{4170} > \frac{1}{\sqrt{4171}} .

Now playing with this example, we see that \frac{1}{\sqrt{k}} > (\frac{1000}{1001})^{4171} for k \in \{4172, 4173,....4179 \} but not for k = 4180 . So c_k = 0 for k \in \{4172,....4179 \} and c_{4180} = 1 . So the first few n_j are \{2, 3, ....4169, 4171, 4180 \} . Of course the gap between the n_j grows as k does.

Now let’s get back to the cartoon example. From this example, we’ll attempt to state a more general result.

Claim: given \sum^{\infty}_{k=1} c_k \frac{1}{k} where c_k = 0 if k contains a 9 as one of its digits, then \sum^{\infty}_{k=1} c_k \frac{1}{k} converges. Hint on how to prove this (without reading the solution): count the number of integers between 10^k and 10^{k+1} that lack a 9 as a digit. Then do a comparison test with a convergent geometric series, noting that every term \frac{1}{10^k}, \frac{1}{10^k + 1}......, \frac{1}{8(10^k) +88} is less than or equal to \frac{1}{10^k} .

How to prove the claim: we can start by “counting” the number of integers between 0 and 10^k 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 8(9) = 72 numbers with no 9 as a digit (hint: consider 10-19, 20-29…80-89) so this means that we have 9 + 8(9) = 9(1+8) = 9^2 numbers between 0 and 99 with no 9 as a digit.

This leads to the conjecture: there are 9^k numbers between 0 and 10^k -1 with no 9 as a digit and (8)9^{k-1} between 10^{k-1} and 10^k-1 with no 9 as a digit.

This is verified by induction. This is true for k = 1

Assume true for k = n . Then to find the number of numbers without a 9 between 10^n and 10^{n+1} -1 we get 8 (9^n) which then means we have 9^n + 8(9^n) = 9^n (8+1) = 9^{n+1} numbers between 0 and 10^{n+1}-1 with no 9 as a digit. So our conjecture is proved by induction.

Now note that 0+ 1 + \frac{1}{2} + ....+ \frac{1}{8} < 8*1*1

\frac{1}{10} + ...+ \frac{1}{18} + \frac{1}{20} + ...+ \frac{1}{28} + \frac{1}{30} + ...+ \frac{1}{88} < 8*9*\frac{1}{10}

\frac{1}{100} + ...\frac{1}{88} + \frac{1}{200} + ....\frac{1}{888} < 8*(9^2)\frac{1}{100}

This establishes that \sum_{k=10^n}^{10^{n+1}-1} c_k \frac{1}{k} < 8*(9^k)\frac{1}{10^k}

So it follows that \sum^{\infty}_{k=1} c_k \frac{1}{k} < 8\sum^{\infty}{k=0} (\frac{9}{10})^k = 8 \frac{1}{1-\frac{9}{10}} = 80 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 \frac{1}{10^k} and \frac{1}{10^{k+1}-1} we allowed 8*(9^k) terms to survive.

This suggests that if we permit up to M (10-\epsilon)^k terms between 10^k and 10^{k+1}-1 to survive (M, \epsilon 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. 🙂


1 Comment »

  1. […] This post is more designed to entertain myself than anything else. This builds up a previous post which talks about deleting enough terms from a divergent series to make it a convergent one. […]

    Pingback by More Fun with Divergent Series: redefining series convergence (Cesàro, etc.) | College Math Teaching — January 26, 2016 @ 10:21 pm

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