<p>No, though coming up with a concrete counterexample is a messy business.</p>
<p>Let <span class="math-container">$y_t=\rho^t$</span>. Then <span class="math-container">$$\sum_{k=0}^\infty \rho^k y_{t+k}=\sum_{k=0}^\infty \rho^{2k+t}=\frac{\rho^t}{1-\rho^2}.$$</span>
Let <span class="math-container">$c_t=1$</span> if <span class="math-container">$t=n^n$</span> for some <span class="math-container">$n$</span> and <span class="math-container">$c_t=0$</span> otherwise. Let <span class="math-container">$t=n^n+1$</span>. Then</p>
<p><span class="math-container">$$\sum_{k=0}^\infty \rho^k c_{t+k}\leq\sum_{k=0}^\infty \rho^{k+(n+1)^{n+1}-n^n-1}=\frac{\rho^{(n+1)^{n+1}-n^n-1}}{1-\rho}&lt;\frac{\rho^{n\cdot n^n}}{1-\rho}=\frac{\rho^{nt}}{1-\rho}.$$</span> So for large <span class="math-container">$n$</span>, we have <span class="math-container">$d_t&gt;0$</span>.</p>

CC BY-SA 4.0
2023-10-08T23:24:25.650Z
2023-10-08T23:24:25.650Z
Michael Greinecker
<p>Yes, I think that <span class="math-container">$d_t$</span> is eventually negative because of the Theorem of Sign Permanence for sequences. I recall it below:</p>
<blockquote>
<p><em><strong>Theorem.</strong></em> Let  <span class="math-container">${a_n}$</span> be a sequence of real numbers. And suppose that <span class="math-container">$\lim_{n\rightarrow \infty} a_n=l&gt;0$</span>,   <span class="math-container">$l\in \mathbb{R}$</span>.</p>
<p>Then exists a <span class="math-container">$n_0 \in \mathbb{N}$</span> such that <span class="math-container">$\forall n&gt;n_0$</span> <span class="math-container">$a_n&gt;0$</span>.</p>
</blockquote>
<p>(The converse if <span class="math-container">$l&lt;0$</span>, of course).</p>
<p>A series of real number is a sequence, the sequence of partial sums, and the sum of a series is the limit of that sequence, so the theorem applies.</p>
<p>Your series converges to a negative number, <span class="math-container">$z/(1-\rho)$</span>, where <span class="math-container">$z$</span> is the limit of <span class="math-container">$(y_{k+k} -c_{t+k})$</span>, so the series is eventually negative.</p>

CC BY-SA 4.0
2022-09-14T10:01:13.523Z
Thanks for the post!  I thought about this further. In my setup, $c_t$ is not necessarily converging to a positive number; it may vacillate arbitrarily close to zero infinitely often (in a way that outpaces $y_t\to 0)$, in which case I think $d_t$ may be negative infinitely often.  But a counterexample would make this concrete.
Golden_Ratio
CC BY-SA 4.0
2022-09-14T10:42:53.023Z
I see, I agree with you. But, then, you don't know if $y_{t+k}-c_{t+k}$ converges, so you can't apply the formula   $$z_t \to z\iff \sum_{k=0}^\infty \rho^k z_{t+k}\to z/(1-\rho).$$ . Moreover, does $d_t$ converge? How can you prove it?
BakerStreet
CC BY-SA 4.0
2022-09-14T10:44:32.737Z
Correct, from the assumptions, $d_t$ need not converge.
Golden_Ratio
CC BY-SA 4.0
2022-09-14T11:19:07.597Z
I think that if $d_t$ doesn't converge, you cant't say nothing about $d_t$ beeing eventually negative. For example, let $\rho=1/2$ and $z=y_{t+k}-c_{t+k}=-2^k$. The series becomes $\sum_{k=0}^\infty (1/2)^k(-2)^k= 1-1+1-1+1....$ , which alternates negative and positive terms.
BakerStreet
CC BY-SA 4.0
2022-09-14T11:22:33.270Z
Perhaps I should clarify. $d_t$ is well-defined so that the series that defines it is a convergent series. However, the sequence $(d_t)_{t\geq 0}$ need not converge.
Golden_Ratio
CC BY-SA 4.0
2022-09-14T11:30:08.080Z
Is see. You mean the sequence $(d_t)$, $t=0,....\infty$, not the series $d_t$.
BakerStreet
CC BY-SA 4.0
2022-09-14T11:59:30.850Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/139188/discussion-between-golden-ratio-and-bakerstreet).
Golden_Ratio
CC BY-SA 4.0
2022-09-13T12:09:24.947Z
2022-09-13T13:54:07.070Z
2022-09-13T13:54:07.070Z
BakerStreet
<p>Suppose <span class="math-container">$d_t=\sum_{k=0}^\infty \rho^k (y_{t+k}-c_{t+k})$</span> and <span class="math-container">$\rho\in (0,1),y_t\geq 0,c_t&gt;0\quad \forall t\geq 0$</span>. Does <span class="math-container">$y_t\to 0,c_t\not\to 0$</span> imply <span class="math-container">$d_t$</span> is eventually negative?</p>
<p>My initial hunch was affirmative, but I want to be sure.</p>
<p>I have already shown that for generic sequence <span class="math-container">$z_t$</span>, we have <span class="math-container">$$z_t \to z\iff \sum_{k=0}^\infty \rho^k z_{t+k}\to z/(1-\rho).$$</span></p>
<p>So <span class="math-container">$c_t\not\to 0$</span> implies <span class="math-container">$\sum_{k=0}^\infty \rho^k c_{t+k}\not\to 0.$</span> Hence <span class="math-container">$\exists \epsilon&gt;0$</span> s.t. <span class="math-container">$\sum_{k=0}^\infty \rho^k c_{t+k}&gt;\epsilon$</span> for infinitely many <span class="math-container">$t$</span>.</p>
<p>Meanwhile, <span class="math-container">$y_t\to 0$</span> implies <span class="math-container">$\sum_{k=0}^\infty \rho^k y_{t+k}\to 0$</span> so for some <span class="math-container">$T$</span>, <span class="math-container">$\sum_{k=0}^\infty \rho^k y_{t+k}&lt;\epsilon$</span> for <span class="math-container">$t\geq T$</span>.</p>
<p>So <span class="math-container">$d_t$</span> should be negative infinitely often but can we say it is eventually always negative?</p>

<mathematical-economics><international-trade>
Is debt eventually negative for nonnegative $y_t\to 0,c_t\not\to 0$?
Golden_Ratio
# Is debt eventually negative for nonnegative $y_t\to 0,c_t\not\to 0$?

(Asked by: username_0 on 2022-09-11)

Suppose $d\_t=\sum\_{k=0}^\infty \rho^k (y\_{t+k}-c\_{t+k})$ and $\rho\in (0,1),y\_t\geq 0,c\_t>0\quad \forall t\geq 0$. Does $y\_t\to 0,c\_t\not\to 0$ imply $d\_t$ is eventually negative?

My initial hunch was affirmative, but I want to be sure.

I have already shown that for generic sequence $z\_t$, we have $$z\_t \to z\iff \sum\_{k=0}^\infty \rho^k z\_{t+k}\to z/(1-\rho).$$

So $c\_t\not\to 0$ implies $\sum\_{k=0}^\infty \rho^k c\_{t+k}\not\to 0.$ Hence $\exists \epsilon>0$ s.t. $\sum\_{k=0}^\infty \rho^k c\_{t+k}>\epsilon$ for infinitely many $t$.

Meanwhile, $y\_t\to 0$ implies $\sum\_{k=0}^\infty \rho^k y\_{t+k}\to 0$ so for some $T$, $\sum\_{k=0}^\infty \rho^k y\_{t+k}<\epsilon$ for $t\geq T$.

So $d\_t$ should be negative infinitely often but can we say it is eventually always negative?

**Tags:** mathematical-economics, international-trade


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## 2 answers

### Answer by username_1 on 2023-10-08. Score: 2

No, though coming up with a concrete counterexample is a messy business.

Let $y\_t=\rho^t$. Then $$\sum\_{k=0}^\infty \rho^k y\_{t+k}=\sum\_{k=0}^\infty \rho^{2k+t}=\frac{\rho^t}{1-\rho^2}.$$
Let $c\_t=1$ if $t=n^n$ for some $n$ and $c\_t=0$ otherwise. Let $t=n^n+1$. Then

$$\sum\_{k=0}^\infty \rho^k c\_{t+k}\leq\sum\_{k=0}^\infty \rho^{k+(n+1)^{n+1}-n^n-1}=\frac{\rho^{(n+1)^{n+1}-n^n-1}}{1-\rho}<\frac{\rho^{n\cdot n^n}}{1-\rho}=\frac{\rho^{nt}}{1-\rho}.$$ So for large $n$, we have $d\_t>0$.


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### Answer by username_2 on 2022-09-13. Score: 0

Yes, I think that $d\_t$ is eventually negative because of the Theorem of Sign Permanence for sequences. I recall it below:

> ***Theorem.*** Let ${a\_n}$ be a sequence of real numbers. And suppose that $\lim\_{n\rightarrow \infty} a\_n=l>0$, $l\in \mathbb{R}$.
>
> Then exists a $n\_0 \in \mathbb{N}$ such that $\forall n>n\_0$ $a\_n>0$.

(The converse if $l<0$, of course).

A series of real number is a sequence, the sequence of partial sums, and the sum of a series is the limit of that sequence, so the theorem applies.

Your series converges to a negative number, $z/(1-\rho)$, where $z$ is the limit of $(y\_{k+k} -c\_{t+k})$, so the series is eventually negative.

> **Comments:**
> **username_0**: Thanks for the post!  I thought about this further. In my setup, $c_t$ is not necessarily converging to a positive number; it may vacillate arbitrarily close to zero infinitely often (in a way that outpaces $y_t\to 0)$, in which case I think $d_t$ may be negative infinitely often.  But a counterexample would make this concrete.  
> **username_2**: I see, I agree with you. But, then, you don't know if $y_{t+k}-c_{t+k}$ converges, so you can't apply the formula   $$z_t \to z\iff \sum_{k=0}^\infty \rho^k z_{t+k}\to z/(1-\rho).$$ . Moreover, does $d_t$ converge? How can you prove it?  
> **username_0**: Correct, from the assumptions, $d_t$ need not converge.  
> **username_2**: I think that if $d_t$ doesn't converge, you cant't say nothing about $d_t$ beeing eventually negative. For example, let $\rho=1/2$ and $z=y_{t+k}-c_{t+k}=-2^k$. The series becomes $\sum_{k=0}^\infty (1/2)^k(-2)^k= 1-1+1-1+1....$ , which alternates negative and positive terms.  
> **username_0**: Perhaps I should clarify. $d_t$ is well-defined so that the series that defines it is a convergent series. However, the sequence $(d_t)_{t\geq 0}$ need not converge.  
> **username_2**: Is see. You mean the sequence $(d_t)$, $t=0,....\infty$, not the series $d_t$.  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/139188/discussion-between-golden-ratio-and-bakerstreet).  

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