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.

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    $\begingroup$ If a series is convergent its terms must tend to $0$. $\endgroup$
    – Gary
    May 15, 2021 at 12:55
  • $\begingroup$ 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. $\endgroup$ May 15, 2021 at 13:59
  • $\begingroup$ 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! $\endgroup$
    – GEdgar
    May 15, 2021 at 15:08
  • $\begingroup$ @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? $\endgroup$ May 15, 2021 at 16:38
  • $\begingroup$ @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. $\endgroup$ May 15, 2021 at 22:10

1 Answer 1


$S_N:= \sum^N R_k$ 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.


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