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

For $k>1$ ($k=2$ in particular for reason) and $\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_{n}}=1$ (it is $a_{n} > 0$ as well, but that is not crucial) we compare these two series:

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

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

Can we claim anything about the convergence of the second knowing the convergence of the first?

You can take if needed be the restrictions: $a_{n+1} > a_{n}$ and even more $a_{n} \in \mathbb{N}$

Of course, we can say that there should be a general requirement that $a_{n+1}$ does not differ in any critical sense from $a_{n}$. For example we may ask that we these two converge

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

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

or something similar that makes certain that we are dealing with convergent series even when split. (Do not take this split convergence as the requirement. This is just an example of possible restrictions. Very good if this one is working. Even better if none is needed.)

This seems plausible, and for any nice example out there it works just fine, but I do not see any direct way of proving it. I suspect that there must be something very wrong with the series if this is not working.

(A normal monotone requirement for alternating series should play no role in the proof as neither of the series needs to be monotone. Only $a_{n}$ can be monotone.)

Let me display the reasoning behind this question. Set $k=2$ in the question as this allows reducing the exponent on its own and it allows to have:

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

Now if we take the condition that for any $N$

$|\sum\limits_{n=1}^{N}(-1)^n (\frac{a_{n+1}}{a_{n}}+1)| < M$

the result might follow if we have something like Dirichlet test just in reverse and that would not require monotone condition or some combination of conditions.

Can this last help to establish maybe the opposite result under some conditions, that if

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

converges then

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

converges?

• The summation over $n\ge1$ can be shifted to $n\ge n_0$ without changing the problem; in particular, $a_n>0$ for all $n$ can be dropped as $a_{n+1}/a_n\to1$ implies that $a_n$ is eventually sign-definite. Denote $x_n=(-1)^n(a_{n+1}/a_n-1)$ and $y_n=(-1)^n((a_{n+1}/a_n)^2-1)$; we have $x_n\to0$ and $y_n\to0$ as (from $a_{n+1}/a_n\to1$), so that $x_n=y_n/2-(-1)^ny_n^2/8+O(y_n^3)$ as $n\to\infty$. Assuming $\sum_{n\ge n_0}y_n$ converges, the divergence of $\sum_{n\ge n_0}x_n$ will potentially require $\sum_{n\ge n_0}(-1)^ny_n^2$ to diverge. – Wadim Zudilin Aug 14 '17 at 13:46
• @WadimZudilin: Cannot fetch last $y_n^2$. Where did 2 come from? – alex.peter Aug 14 '17 at 13:56
• I have already given a solution in the yesterday comment: take, e.g., $y_{2n-1}=0$ (that is, $a_{2n}=a_{2n-1}$) and $y_{2n}=(-1)^n/\sqrt{n}$ (that is, $(a_{2n+1}/a_{2n})^2=1+(-1)^n/\sqrt{n}$). Then $\sum_{n=1}^\infty y_n^k$ converges for any $k\ge1$ different from 2, in particular, for $k=1$ and 3, and diverges for $k=2$. Thus $\sum_{n=1}^\infty x_n$ diverges. The connection between $x_n$ and $y_n$ (was that your question?) comes from solving $x(2+x)=y$. – Wadim Zudilin Aug 15 '17 at 9:25

No, one can create sequences in which the first sequence converges but the second does not (or vice versa).

For sake of argument take $k=2$. To begin with let us ignore the requirement that the $a_n$ be natural numbers. Let $\varepsilon_m>0$ be a sequence of numbers tending very slowly to zero (e.g. $\varepsilon_m = 1 / \log \log(m+100)$). Define $a_n$ so that the ratios take the form $$\frac{a_{4m+1}}{a_{4m}} = (1 + \varepsilon_m)^{1/2}$$ $$\frac{a_{4m+2}}{a_{4m+1}} = (1 + 2\varepsilon_m)^{1/2}$$ $$\frac{a_{4m+3}}{a_{4m+2}} = (1 + 3\varepsilon_m)^{1/2}$$ $$\frac{a_{4m+4}}{a_{4m+3}} = (1 + 2\varepsilon_m)^{1/2}$$ (setting $a_1=1$ say). Then $\sum_n (-1)^n ((\frac{a_{n+1}}{a_n})^2 - 1)$ converges (indeed the series is designed so that the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ vanish, and the summands go to zero). On the other hand, from Taylor expansion, the sum of four consecutive terms $n = 4m, 4m+1,4m+2,4m+3$ of $\sum_n (-1)^n (\frac{a_{n+1}}{a_n} - 1)$ sums to $-\frac{1}{4} \varepsilon_m^2 + O(\varepsilon_m^3)$ if I did the arithmetic correctly, and this will diverge if $\varepsilon_m$ decays slowly enough.

To make the $a_n$ integer, one should take the real sequence constructed above and replace each element by its integer part. The sequence grows almost exponentially fast, so one can check that the errors incurred by doing so do not disrupt the convergence or divergence.

One can concoct similar examples for other $k$, or assuming the convergence or divergence of the other series mentioned in the post. More generally, once the summands are allowed to oscillate, the (conditional) convergence of one series says almost nothing about the convergence of any other series due to the lack of any useful comparison inequalities for oscillating sums.

This is false, and the reason for this is fairly general, so in fact pretty much any such statement has to be false. Let's write $b_n=\frac{a_{n+1}}{a_n}-1$, so $b_n\to 0$, and we're now asking if the convergence of $\sum (-1)^n b_n(2+b_n)$ implies that of $\sum (-1)^n b_n$.

This will not be the case if the convergence of the first sum depended on near perfect cancellations between potentially large contributions. Here's a concrete example: Let $f(x)=-1+\sqrt{1+x}$; notice that $f(x)>0$ and $f(x)(2+f(x))=x$. Now take $b_1=f(1/N)$, $b_n=f(1/N^2)$ for $n=2,4,6, \ldots , 2N$, and $b_n=0$ for the odd $n>1$ from this interval.

Define the whole sequence $b_n$ by a succession of intervals $I_k$ of this type, with $N_k\to\infty$ (to be chosen later). Then $\sum (-1)^nb_n(2+b_n)$ converges (to $0$). However, since $f(x)=x/2 - x^2/8+O(x^3)$ for small $x$, we have that $$8\sum_{n\in I_k} (-1)^n b_n =\frac{1}{N_k^2} + O(1/N_k^3) .$$ So $\sum (-1)^n b_n$ will diverge if we take $N_k$ such that $\sum 1/N_k^2=\infty$.

(The converse is also false, and in fact this is what I did in the first, now edited away, version of this answer because I had misread the question.)