# Switching terms in a series - theorem

estro
Suppose a_n defined in the following way:
$$b_{2n}=a_{2n-1}$$
$$b_{2n-1}=a_{2n}$$
I know that $$\sum a_n$$ is convergent.

This how I proved that $$\sum b_n$$ is also convergent.

$$S_k=\sum_{n=1}^k b_n = \sum_{n=1}^k b_{2n} + \sum_{n=1}^k b_{2n-1} = \sum_{n=1}^k a_{2n-1} + \sum_{n=1}^k a_{2n} = \sum_{n=1}^k a_n \leq M$$

Am I right?

Homework Helper
So you are saying that $a_1+ a_2+ a_3+ a_4+ a_5+ a_6+ ...$ becomes $a_2+ a_1+ a_4+ a_3+ a_6+ a_5+ ...$?

What you have proved is that the sequence of partial sums is bounded above. If those partial sums are not increasing (if the $a_n$ are not all non-negative) that does not prove convergence.

estro
Thanks, I now understand my mistake, but I can't find contra-example for this theorem.
My intuition locks me into thinking that switching 2 closest terms won't change convergence.
How should I approach such problems?
I will try thinking about Cauchy Criterion.

estro
$$\sum a_n \rightarrow L\ \Rightarrow\ a_n\rightarrow0\ \Rightarrow\ \forall\ n>N_1\ a_n<\epsilon/4$$
$$\sum a_n \rightarrow L\ \Rightarrow\ \forall\ m,n>N_2 |\\ \sum_{n+1}^m a_n|\leq \epsilon/4$$

So for all n>max{N_1+1,N_2+1}
$$|\sum_{k=n+1}^m b_k| \leq |a_n + \sum_{k=n+1}^m a_k + a_{m+1}| \leq \epsilon$$

Is this idea right?

Last edited:
estro
I also try to explain in words my idea as I getting hard time with latex:

Because $$\sum a_n$$ is convergent and thanks to Cauchy Criterion we know that after some n I can sum a_n as many times as I want while keep the sum small as I want.
So I took little more terms to express $$\sum b_n$$ and to meet the Cauchy Criterion again.
I hope I expressed myself clearly

estro