Is a strictly increasing necessarily a subsequent of an arithmetic progression?

Question

Let $(b_n)_{n\in\mathbb{N}}$ be a strictly increasing sequence such that, $$\lim_{n\to+\infty} b_n - b_{n-1} > 0$$ Is there an arithmetic progression whose subsequent is $(b_n)_{n\in\mathbb{N}}$?

Update By $\lim_{n\to+\infty} b_n - b_{n-1} > 0$, I mean that the limit is not equal to 0, but it is possible that $\lim_{n\to+\infty} b_n - b_{n-1} = +\infty$

Answer 1

Suppose $(b_n)_n$ is a subsequence of an arithmetic progression $(a_n)_n$. Let $d := a_2 - a_1$ be the common difference of the latter. Since $(b_n)_n$ is strictly increasing (by your hypothesis), it follows that $d > 0$. Moreover, $$c_n :=\frac{b_{n + 1} - b_n}{d}$$ is a positive integer for all $n$.

Now, by your hypothesis about the $\lim_n [b_n - b_{n - 1}]$ existing, we see that $(c_n)_n$ converges. However, $c_n$ is integer-valued and thus, must be eventually constant. (Edit: Here I've been assuming that the limit is finite.)
In turn, the difference $b_{n + 1} - b_n$ must eventually be constant.

However, we now see that the comment by dvix gives a counterexample. Indeed, taking $$b_{n + 1} = b_n + 1 + \frac{1}{2^n},$$ we see that

• $(b_n)_n$ is strictly increasing,
• $\lim_{n \to \infty} (b_n - b_{n - 1}) = 1> 0$, but
• $b_{n + 1} - b_n = 1 + 2^{-n}$ is not eventually constant.

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Edit 2. Even if $\lim_{n \to \infty}[b_n - b_{n - 1}] = \infty$, it is not necessary that $(b_n)_n$ is a subsequence of an arithmetic sequence.
Consider $b_n = b_{n - 1} + n + 2^{-n}$.

The proof is essentially to show that you cannot find any $d > 0$ such that $$\frac{b_{n + 1} - b_n}{d}$$ is an integer for all $n$.

Answer 2

No in general. An increasing arithmetic sequence has its difference between two consecutive terms equal to $r \gt 0$.

And for a general increasing sequence, the difference $b_n -b_{n-1}$ can be strictly positive and converge to zero.