Can a telescopic series with $a_n$ not tending to zero converge?

Question

Let $\{a_n\}$ be a sequence with $a_n\to l\ne0$ as $n\to\infty$.

Then, does the series $$\sum^\infty_{n=1}(a_n-a_{n+1})$$ necessarily converge?

I don’t know which of the following arguments is correct:

The series converges:

Because $$\sum^N_{n=1}(a_n-a_{n+1})=a_1-a_N$$ Taking limit $N\to\infty$ on both sides, the series converges to $a_1-l$.

The series diverges:

Set $a_n=1$ for all $n$.

Then the series is equivalent to $$1-1+1-1+1-1\cdots$$ which does not converge.

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Furthermore, if the series does not always converge, under what conditions would the series converge?

Answer 1

You are assuming we don't know how the series is structured. But we do - instead of having $$1-1+1-1+\cdots$$ we instead have $$(1-1)+(1-1)+\cdots$$ so every term cancels and the series clearly converges! This is more evident from the sigma notation, since $a_n = a_{n+1} = 1$ so $\sum (a_n - a_{n+1}) = \sum 0 = 0$

Answer 2

You must realize that the infinite series (which is rigorously denoted by a summation expression) is defined as the limit of the partial sums if it exists, and the partial sums are in turn defined in a certain precise way. One is not permitted to do any rearrangement, especially by 'intuition', unless one proves that doing so does not change the limit of the partial sums. This is why $\sum_{k=0}^\infty (-1)^k$ diverges while $\sum_{k=0}^\infty ((-1)^{2k}+(-1)^{2k+1})$ converges.