example of a rational series, conditionally convergent, such that it converges to a rational number

Question

Is there any example of a rational series (of rational terms), conditionally convergent (that is, not absolutely convergent) such that it converges to a rational number?

For example the Leibniz formula for π/4 is an example, but it converges to an irrational number. Thanks a lot.

Answer 1

Let $a_n = \left(-\frac{1}{2}\right)^n,\ n\geq 0.\quad \displaystyle\sum_{n=0}^{\infty} a_n $ is a geometric series, with $\ \displaystyle\sum_{n=0}^{\infty} a_n = \frac{1}{1-\left(-\frac{1}{2}\right)} = \frac{2}{3}.$

Let $b_0 = 1,\ b_1 = -1,\ b_2 = \frac{1}{2},\ b_3 = -\frac{1}{2},\ b_4 = \frac{1}{3},\ b_5 = -\frac{1}{3},\ldots. $

Inter-weave the two sequences to get:

$$a_0,\ b_0,\ a_1,\ b_1,\ a_2,\ b_2,\ \ldots.\ $$

Summing this sequence in this order gives:

$$a_0 + b_0 + a_1 + b_1 + a_2 + b_2 + \ldots = \frac{2}{3}. $$

And since $(b_n)_{n\in\mathbb{N}}\ $ is a subsequence of the sequence $a_0,\ b_0,\ a_1,\ b_1,\ a_2,\ b_2,\ \ldots,\ $ it is clear that the above series is conditionally convergent.

Answer 2

You can use the Riemann rearrangement theorem, for example with the alternated harmonic series. For every positive term (the reciprocal of an odd integer), take four negative terms (reciprocals of even integers) : \begin{align*} 1 &- 1/2 - 1/4 - 1/6 - 1/8\\ + 1/3 &- 1/10 - 1/12 - 1/14 - 1/16\\ + 1/5 &- 1/18 - 1/20 - 1/22 - 1/24\\ + 1/7 &- \ldots \end{align*} This is clearly conditionally convergent since it is a rearrangement of the alternated harmonic series. Now, the first $n$ rows sum to : \begin{align*} \sum_{k = 1}^{n} \frac{1}{2k-1} - \sum_{k = 1}^{4n}\frac{1}{2k} &= \left(H_{2n} - \frac{H_n}{2}\right) - \frac{H_{4n}}{2}\\ &= \ln(2n) + \gamma - \frac{\ln(n) + \gamma}{2} - \frac{\ln(4n) + \gamma}{2} + o(1)\\ &= \ln(2) + \ln(n) - \frac{\ln(n)}{2} - \frac{\ln(4)}{2} - \frac{\ln(n)}{2} + o(1)\\ &= o(1) \end{align*} Whence the sum is zero.