$\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}}{a_{n}}-1)$ converges. Does $\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+1}}{a_{n}}-1)$ converge?

Question

The question formulated as

If $\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}}{a_{n}}-1)$ converges does $\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+1}}{a_{n}}-1)$ converge?

is asking if we can reduce the counter under these assumptions:

1. $a_{n+1} > a_{n}$

2. $a_{n} \in \mathbb{N}$

3. $\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_{n}}=1$

It seems to me that the answer is generally no, but I cannot find any decisive example that would make it not working. Even more I cannot find the condition when it does work.

Reasoning: $\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}}{a_{n}}-1)=$

$=\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}}{a_{n}}-\frac{a_{n+1}}{a_{n}}+\frac{a_{n+1}}{a_{n}}-1)$

$=\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}-a_{n+1}}{a_{n}})+\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+1}}{a_{n}}-1)$

Now we need this to have limit and we are done:

$\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}-a_{n+1}}{a_{n}})$

Although it might be possible that we can prove that this is bounded, generally this can oscillate. Or?

Answer 1

The following sequence provides a counterexample and satisfies assumptions 1,2 and 3: $$a_{2n+1}=(n+1)(n+2), a_{2n}=n(n+1)+1=a_{2n-1}+1.$$

To check this, let

$$b_n=(-1)^n(\frac{a_{n+2}}{a_n}-1), c_n=(-1)^n(\frac{a_{n+1}}{a_n}-1).$$

Then your sums are equal to

$$\sum_{n=1}^{+\infty} b_n \text{ and } \sum_{n=0}^{+\infty} c_n.$$

But we have

$$b_{2n+1}=-(\frac{n+3}{n+1}-1)=-\frac{2}{n+1}=-\frac{2}{n}+O(1/n^2),$$

$$b_{2n}=\frac{(n+1)(n+2)+1}{n(n+1)+1}-1=\frac{2n+2}{n(n+1)+1}=\frac{2}{n}+O(1/n^2),$$

therefore the first series is convergent.

On the other hand,

$$c_{2n+1}=-(\frac{(n+1)(n+2)+1}{(n+1)(n+2)}-1)=O(1/n^2)$$

and

$$c_{2n}=(\frac{(n+1)(n+2)}{n(n+1)+1}-1)=\frac{2n}{n^2+n+1}=2/n+O(1/n^2),$$

hence the second sum is divergent.