2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$
    – tst
    Nov 10 '14 at 19:50
1
$\begingroup$

Since the general term does not go to zero, the series never converges.

$\endgroup$
1
  • $\begingroup$ You're right, it has been a while for me. Forgot about this fundamental criteria. $\endgroup$
    – Patrick
    Nov 10 '14 at 19:54
0
$\begingroup$

I think this can interest you. I'd add that you can use another definition of "convergence of series".

Yes, this one does not converge. But, as you pointed out with your example, you can think of it as a "mean" of its term.

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!).

You can call it "Cesaro-convergence" or C-convergence (see this).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.