Is every series a telescoping series?

This question may seem silly at first. We say that a series $$\sum a_n$$ is a telescoping series if there exists a sequence $$(b_n)$$ with $$a_n=b_n-b_{n+1}$$ for every $$n$$. One can show that $$\sum a_n$$ converges if and only if the sequence $$(b_n)$$ converges. It is thus desirable to know if a series is a telescoping series, and then find a "telescoping form".

This begs the question, which is in the title, is every series a telescoping series? I have little doubt that the answer is that not every series is a telescoping series. The problem I have in finding a counterexample is that it seems hard to prove that given a sequence $$(a_n)$$ there is no sequence $$(b_n)$$ such that $$a_n=b_n-b_{n+1}$$ for every $$n\in\mathbb{N}$$.

I have another question which is related enough to the first question so I'll ask it in this post. If we have $$a_n=b_n-b_{n+1}$$ then we also have $$a_n=c_n-c_{n+1}$$ where $$c_n=b_n+c$$ for some $$c$$. Somewhat analogous to antiderivatives, given sequences $$(a_n),(b_n)$$ with $$a_n=b_n-b_{n+1}$$ then if $$(c_n)$$ satisfies $$a_n=c_n-c_{n+1}$$ then does there exist a $$c$$ such that $$c_n=b_n+c$$? If the answer is affirmative then given a telescoping series we can find all of its "telescoping forms".

• Even for a telescoping series you would want $\lim_{n\to\infty} b_n = 0$. This would rule out the existence of a $c\neq 0$, suggested in the second part of your question. Commented Oct 4, 2023 at 7:37
• I think most people, when they talk about a telescoping series in practice, they mean a usefully telescoping series. Commented Oct 4, 2023 at 7:46
• Usually in telescoping series, $a_n = b_n-b_{n-1}$, not $b_{n+1}$. With this correction, let $b_n = \sum_{i=0}^n a_i$. Then $a_n = b_n - b_{n-1}$. For your direction $b_n = -\sum_{i=0}^{n-1} a_i$ works. Commented Oct 4, 2023 at 23:17

Part 1 (ignoring convergence): If you don't care about convergence or "meaningful" series:

1. Given the sequence $$(a_n)$$, you can always find the terms of the telescoping series from the recursion $$b_{n+1} = a_n + b_n$$, for any $$b_0$$ that can be chosen freely.
2. Taking the difference between the recursions for $$b_n$$ and $$c_n$$, you'll find that $$b_{n+1} - c_{n+1} = b_n - c_n$$, i.e. the difference between the $$b_n$$ and $$c_n$$ is independent of $$n$$, and we denote it by $$c$$ in your notation. Note that $$c = b_0 - c_0$$, i.e. it is specified by the initial conditions.

Part 2 (convergent series): Does it still work if we require the appearing series to be convergent? The answer is yes, and the telescopic sum is unique.

Assume that $$\sum a_n$$ is convergent, and require $$\lim_{n\to\infty} b_n= 0$$, so that the telesopic series may be convergent as well (necessary condition).

Define $$b_n = a_0 + a_1 +\dots + a_{n-1} - \sum_{j=0}^\infty a_j$$ and $$b_0 =-\sum_{j=0}^\infty a_j$$. This will have the property that $$\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty (b_{n+1} - b_n),$$ and $$\lim_{n\to\infty} b_n = 0$$, to ensure your telescopic series is convergent as well. Hence, for every convergent series, there exists a corresponding telescoping series that converges.

Uniqueness follows from Part 1.2: If there were a different telescopic series with terms $$(c_n)$$, there would be a nonzero $$c$$ such that $$b_n - c_n = c$$. However, then $$c_n\to c\neq 0$$, which violates our assumption.

• Why the downvote? Commented Oct 4, 2023 at 7:48
• Good question! Upvote from me. Commented Oct 4, 2023 at 7:50
• Shouldn´t $b_0=-\sum_{j=0}^{\infty}a_j$? Commented Oct 4, 2023 at 7:55
• Thanks for the response, but this doesn't answer the questions. Does there exist a divergent series that is not a telescoping series? Ans also the second question hasn't been answered. Commented Oct 4, 2023 at 8:21
• The downvote may be because you buried the lede: "Hence, for every convergent series, there exists a corresponding telescoping series that converges." I had to read through twice to figure out what your answer was saying. It may also help to follow up on Arthur's concept of a "usefully" telescoping series, although that it harder to quantify. Commented Oct 4, 2023 at 15:58