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

Question

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?

Answer 1

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 $\sum\limits_{k=1}^{n} f(k) \to \infty$

Answer 2

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.