# If $\sum_{i =1}^ja_i$ is divergent and $a_n \sim f(n)$, is it true that $\lim_{k \to \infty}\sum_{i =1}^ka_i = \lim_{k \to \infty}\sum_{i =1}^kf(i)$?

Let $$a_n$$ be a infinite monotonic sequence which tends toward infinity or minus infinity, and let $$a_k \sim f(k)$$. Then, I ask whether the following is true:

$$\lim_{j \to \infty}\sum_{i=1}^ja_i=\lim_{j \to \infty}\sum_{i=1}^jf(i)$$

Is the same true for any $$a_n$$ such that $$\lim_{j \to \infty}\sum_{i =1}^ja_i$$ is divergent? I want to say that the above claim is true purely by intuition, but would like to know how one might go about rigorously proving it. Would it follow the argument that while some arbitrary number of terms of $$a_n$$ may not be at all accurately approximated by $$f(n)$$, as $$n \to \infty$$ we have an infinite number of terms for which the relative error approximation by $$f$$ is approaching zero, and then those initial terms must be negligible?

• In both cases, $\lim_{j \to \infty}\sum_{i=1}^ja_i$, or in other words, $\sum_{i=1}^\infty a_i$ is not convergent. Be very careful proving or using equations where one side is not a sensible number. – Theo Bendit Jul 22 '19 at 5:20
• (FWIW, frequently in calculus one writes limits equal to $\pm\infty$ sensibly.) – runway44 Jul 22 '19 at 5:21
• @TheoBendit yes! But this question follows something which I had seen in a proof which I questioned; what I ask here is a more general analog of it. Perhaps it would be better of me to include it in my question... – Descartes Before the Horse Jul 22 '19 at 5:23
• @heepo I think that would probably be a good idea, though you seem to have some good answers. I personally tend not to like using $\infty$ in equations, as it typically is used in a way that obscures a separate, special case proof of whatever is being proven. There are some exceptions though. – Theo Bendit Jul 22 '19 at 5:27
• Actually something much stronger holds: as $k\to \infty$ one has the asymptotics $$\sum_{i =1}^ka_i \sim\sum_{i =1}^kf(i)$$ – Gabriel Romon Jul 22 '19 at 8:20

Consider the case where the limit is $$\infty$$. [A similar argument holds when the limit is $$-\infty$$]. If $$\frac {a_n} {f(n)} \to 1$$ then there exists $$n_0$$ such that $$\frac 1 2 \sum\limits_{k=n_0}^{n} a_k \leq \sum\limits_{k=n_0}^{n} f(k) \leq 2\sum\limits_{k=n_0}^{n} a_k$$ so $$\sum\limits_{k=1}^{n} a_k \to \infty$$ iff $$\sum\limits_{k=1}^{n} f(k) \to \infty$$
• Does this also mean the sum of $a_i$ is asymptotic to the sum of $f(i)$? Sorry for formatting, I am on mobile. – Descartes Before the Horse Jul 22 '19 at 15:38
WLOG assume $$a_n\to+\infty$$. Pick $$0<\varepsilon<1$$ and $$N$$ for which $$n\ge N$$ implies $$f(n) \ge (1-\varepsilon)a_n$$. (We can do this because the ratio $$f(n)/a_n$$ is eventually trapped to within $$\varepsilon$$ of $$1$$.) Then for $$k\ge N$$,
$$\sum_{n=1}^k f(n)\ge \sum_{n=1}^{N-1} f(n)+(1-\varepsilon)\sum_{n=N}^{k} a_n.$$
The sum $$\displaystyle \sum_{n=1}^{N-1}f(n)$$ is constant and $$\displaystyle \sum_{n=N}^{k} a_n$$ diverges with $$k$$, thus $$\displaystyle \sum_{n=1}^n f(n)$$ diverges.