# If $\sum\limits_{t=1}^{\infty}R_t$ is finite with $R_t\geq 0$, does $\lim_{t\rightarrow \infty}R_t=0$?

This question arises from my study of economic models with an infinite time horizon in which production is constrained by a finite initial stock of a non-renewable resource.

Given $$\sum\limits_{t=1}^{\infty}R_t = S$$ where $$S$$ is finite and $$R_t\geq 0\,\forall t$$, can it be inferred that $$\lim_{t\rightarrow \infty}R_t=0$$, and if so how can this be proved?

My thoughts: From the definition of a limit and given $$R_t\geq 0$$, we require that for any $$\epsilon >0$$, there exists $$k \in \mathbf{N}$$ such that $$R_t<\epsilon\, \forall t>k$$. If $$R_t$$ decreases monotonically it seems obvious that this must hold. However, the fact that the sum is finite does not require $$R_t$$ to decrease monotonically. It could be that $$R_t=0$$ for most $$t$$ and is positive only for occasional $$t$$ at widely spaced intervals. In that case the result doesn't seem obvious.

• If a series is convergent its terms must tend to $0$.
– Gary
May 15, 2021 at 12:55
• To be fair, this is true only if you are summing with a Cauchy method. If you were to use Cesaro criterion to sum, this result wouldn't hold. May 15, 2021 at 13:59
• This fact (without the redundant assumption $R_t \ge 0$) should be in every beginning calculus textbook. Congratulations if you discovered it without ever taking such a course! May 15, 2021 at 15:08
• @DavideTrono In the application I am interested in, $R_t$ is the quantity of a resource used in period $t$. Is there any reason why the Cauchy method - which to me seems much more natural than the Cesaro method - should not be used to obtain the sum to infinity? May 15, 2021 at 16:38
• @AdamBailey I don't think so. But if one day you will end up with a similar concept (infinite sum of resources) which, by any unluck, diverges, then maybe using stronger methods than Cauchys one could be useful. May 15, 2021 at 22:10

$$S_N:= \sum^N R_k$$ converges, so is a Cauchy sequence, so $$|\sum_{k=1}^{n+1}R_k-\sum_{k=1}^{n}R_k|= |R_{n+1}|$$ becomes arbitrarily small.
This also shows that the assumption $$R_k\ge 0$$ can be omitted.