# Convergence of $\sum_{n=0}^{\infty}(-1)^{a_n}$ for non-negative integer $a_n$.

I am interested (purely for the sake of curiosity) in sums of the form $\sum_{n\geq0}(-1)^{a_n}$ and particularly which sequences $a_n$ (of non-negative integers only) lead to the series being conditionally convergent (I say conditionally because clearly $\sum|(-1)^{a_n}|$ diverges). We can reduce each such sequence of exponents modulo $2$ since the parity completely determines the value of each summand.

My question is: "Does there exist a non-trivial condition on the $a_n$ that is both necessary and sufficient for such a series to converge?"

A motivating example: Does the series $S = \sum_{n=0}^{\infty}(-1)^{F_n}$ , where $F_n$ is the $n$-th Fibonacci number, converge?

$$S = 1 - 1 - 1 + 1 - 1 - 1 + \cdots$$

$$= (1 - 1) - 1 + (1 - 1) - 1 + \cdots$$

$$= 0 - 1 + 0 - 1 + \cdots = -\infty$$

But the above may just be a consequence of Riemanns rearrangement theorem if the series is in fact conditionally convergent.

• Since the absolute value of the terms is 1, the series does not converge in the "traditional" sense. It may converge if you sum it up in a special way, but there are complications. I would suggest you to look at this wikipedia article en.wikipedia.org/wiki/Summation_of_Grandi%27s_series
– tst
Commented Nov 10, 2014 at 19:50

Let's define $$S_n = \sum_{k=0}^n (-1)^{a_k}$$. What can you say about the convergence of $$u_n := \dfrac{S_0 + ... + S_n}{n+1}$$? It depends on $$a_n$$. One could say that it defines another way of convergence of the serie (even if stricto sensu, it does NOT converges with its classic definition!).