Property of a divergent sequence $\{a_n\}$ of positive real numbers satisfying $\lim_{n\to\infty}(a_{n+1}-a_n)=0$

Question

Let $\{a_n\}$ be an unbounded sequence of positive real numbers s. t. $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$.

Question:

Is $a_n$ necessarily equal to a partial sum $\sum_{k=1}^n b_k$ of some sequence $\{b_k\}$ s. t. $\lim\limits_{k\to\infty}b_k=0$, but $\sum_{k=1}^\infty b_k=\infty$?

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Thoughts:

Since $\{a_n\}$ is an unbounded sequence of positive real numbers, I thought it must diverge.

On the other hand, $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$ reminded me of the limits of the form $\lim\limits_{n\to\infty}\frac{x_n}{y_n}$, where also $x_n\to\infty$, which could be evaluated via Stolz-Cesaro, but the only examples of divergent sequences $\{a_n\}$ satisfying $\lim\limits_{n\to\infty}(a_{n+1}-a_n)=0$ I could think of were the partial sums $\sum_{k=1}^nb_k$ of a sequence $b_k$ converging to $0$.

However, my statement seems fragile and I'm pretty sceptical.

Is there any counter-example?

I also took into account sequences that contain infinite products, but, if the logarithms are involved, we could transform the product into a sum.

Thank you in advance!

Answer 1

As already stated in the comment, if $a_n=\sum_{k=1}^n{b_k}$ then $b_n=a_n-a_{n-1}$ is the only possibility. Therefore, you have $\lim\limits_{k\to\infty}b_k=\lim\limits_{k\to\infty}a_k-a_{k-1}=0$.

The second part is more problematic since $\sum_{k=1}^\infty{b_k} = \infty \iff \lim\limits_{k\to\infty}a_k=\infty$ which is not guaranteed. Check the following counter example: $$a_n=\ln(n)(1+\sin(\sqrt{n})), n\ge1, a_0=0$$

Answer 2

I later saw a bit trivial example of a divergent positive sequence $\{a_n\}$ with the property $\lim\limits_{n\to\infty}(a_{n+1}-a_n)$ worth mentioning: $a_n=\sqrt n$. $$\lim_{n\to\infty}a_n=+\infty\quad\land\quad \lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}(\sqrt{n+1}-\sqrt n)=\lim_{n\to\infty}\frac1{\sqrt{n+1}+\sqrt n}=0$$