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.

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