College Math Teaching

February 5, 2016

More fun with selective sums of divergent series

Just a reminder: if \sum_{k=1}^{\infty} a_k is a series and c_1, c_2, ...c_n ,, is some sequence consisting of 0’s and 1’s then a selective sum of the series is \sum_{k=1}^{\infty} c_k a_k . The selective sum concept is discussed in the MAA book Real Infinite Series (MAA Textbooks) by Bonar and Khoury (2006) and I was introduced to the concept by Ferdinands’s article Selective Sums of an Infinite Series in the June 2015 edition of Mathematics Magazine (Vol. 88, 179-185).

There is much of interest there, especially if one considers convergent series or alternating series.

This post will be about divergent series of positive terms for which lim_{n \rightarrow \infty} a_n = 0 and a_{n+1} < a_n for all n .

The first fun result is this one: any selected x > 0 is a selective sum of such a series. The proof of this isn’t that bad. Since lim_{n \rightarrow \infty} a_n = 0 we can find a smallest n such that a_n \leq x . Clearly if a_n = x we are done: our selective sum has c_n = 1 and the rest of the c_k = 0 .

If not, set n_1 = n and note that because the series diverges, there is a largest m_1 so that \sum_{k=n_1}^{m_1} a_k \leq x . Now if \sum_{k=n_1}^{m_1} a_k = x we are done, else let \epsilon_1 = x - \sum_{k=n_1}^{m_1} a_k and note \epsilon_1 < a_{m_1+1} . Now because the a_k tend to zero, there is some first n_2 so that a_{n_2} \leq \epsilon_1 . If this is equality then the required sum is a_{n_2} + \sum_{k=n_1}^{m_1} a_k , else we can find the largest m_2 so that \sum_{k=n_1}^{m_1} a_k + \sum_{k=n_2}^{m_2} a_k \leq x

This procedure can be continued indefinitely. So if we label \sum_{k=n_j}^{m_{j}} a_k = s_j we see that s_1 + s_2 + ...s_{n} = t_{n} form an increasing, bounded sequence which converges to the least upper bound of its range, and it isn’t hard to see that the least upper bound is x because x-t_{n} =\epsilon_n < a_{m_n+1}

So now that we can obtain any positive real number as the selective sum of such a series, what can we say about the set of all selective sums for which almost all of the c_k = 0 (that is, all but a finite number of the c_k are zero).

Answer: the set of all such selective sums are dense in the real line, and this isn’t that hard to see, given our above construction. Let (a,b) be any open interval in the real line and let a < x < b . Then one can find some N such that for all n > N we have x - a_n > a . Now consider our construction and choose m large enough such that x - t_m > x - a_n > a . Then the t_m represents the finite selected sum that lies in the interval (a,b) .

We can be even more specific if we now look at a specific series, such as the harmonic series \sum_{k=1}^{\infty} \frac{1}{k} . We know that the set of finite selected sums forms a dense subset of the real line. But it turns out that the set of select sums is the rationals. I’ll give a slightly different proof than one finds in Bonar and Khoury.

First we prove that every rational in (0,1] is a finite select sum. Clearly 1 is a finite select sum. Otherwise: Given \frac{p}{q} we can find the minimum n so that \frac{1}{n} \leq \frac{p}{q} < \frac{1}{n-1} . If \frac{p}{q} = \frac{1}{n} we are done. Otherwise: the strict inequality shows that pn-p < q which means pn-q < p . Then note \frac{p}{q} - \frac{1}{n} = \frac{pn-q}{qn} and this fraction has a strictly smaller numerator than p . So we can repeat our process with this new rational number. And this process must eventually terminate because the numerators generated from this process form a strictly decreasing sequence of positive integers. The process can only terminate when the new faction has a numerator of 1. Hence the original fraction is some sum of fractions with numerator 1.

Now if the rational number r in question is greater than one, one finds n_1 so that \sum^{n_1}_{k=1} \frac{1}{k} \leq r but \sum^{n_1+1}_{k=1} \frac{1}{k} > r . Then write r-\sum^{n_1+1}_{k=1} \frac{1}{k} and note that its magnitude is less than \frac{1}{n_1+1} . We then use the procedure for numbers in (0,1) noting that our starting point excludes the previously used terms of the harmonic series.

There is more we can do, but I’ll stop here for now.

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
harmonic-series

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. 🙂

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