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".

  • $\begingroup$ 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. $\endgroup$
    – DominikS
    Commented Oct 4, 2023 at 7:37
  • 7
    $\begingroup$ I think most people, when they talk about a telescoping series in practice, they mean a usefully telescoping series. $\endgroup$
    – Arthur
    Commented Oct 4, 2023 at 7:46
  • $\begingroup$ 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. $\endgroup$ Commented Oct 4, 2023 at 23:17

1 Answer 1


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.

  • 2
    $\begingroup$ Why the downvote? $\endgroup$
    – DominikS
    Commented Oct 4, 2023 at 7:48
  • $\begingroup$ Good question! Upvote from me. $\endgroup$ Commented Oct 4, 2023 at 7:50
  • 1
    $\begingroup$ Shouldn´t $b_0=-\sum_{j=0}^{\infty}a_j$? $\endgroup$ Commented Oct 4, 2023 at 7:55
  • 1
    $\begingroup$ 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. $\endgroup$
    – Hilbert
    Commented Oct 4, 2023 at 8:21
  • 1
    $\begingroup$ 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. $\endgroup$
    – Teepeemm
    Commented Oct 4, 2023 at 15:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .