Why isn't $\sum_{n=1}^{\infty}(-1)^n$ a telescoping series?

Question

I read that the reason $\sum_{n=0}^\infty(-1)^n$ does not converge is because the terms alternate between -1 and 0. So I realize that this sentence does answer my question, but what I'm having difficulty understanding is that why wouldn't it collapse to 0 at infinity in the same way that another alternating series, even if it doesn't collapse to 0 before infinity, does?

Answer 1

Consider the sequence $(a_n)_n$ defined as $a_n = (-1)^n.$

Now you can form the sequence of the partial sums, $(s_n)_n$, as

$$s_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n.$$

By definition, $\sum_{k=1}^{\infty} a_n = \lim s_n.$

Let’s study the convergence of $(s_n)_{n}.$ Consider this two subsequences of $(s_{n})_{n}.$

$$s_{2n} = (-1)^{2n} = 1 \quad \quad s_{2n-1} = (-1)^{2n-1} = -1$$

As you can easily check

$$\lim s_{2n} = 1 \neq -1 = \lim s_{2n-1}.$$

So there are two subsequences of $s_n$ that converge to different values. Therefore $s_n$ does not converge.

If a sequence converges, then every subsequence converges for the same limit.

Check this for more information on this result

Prove: If a sequence converges, then every subsequence converges to the same limit.

Therefore $\lim s_n$ does not exist. Hence $\sum_{k=1}^{\infty}$ diverges.

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Note that in this case the cancellation of the terms in $s_n$ wouldn’t work because the result would depend on what $n$ you are deciding. That’s why this is not a telescoping series, you can’t express it as a certain amount of terms after cancellation and still have the same result

Answer 2

A telescoping series (sum) is usually defined as a finite series in which consecutive terms cancel, so that you are only left with the initial and final terms. The series you're considering is an infinite series. In general, you can't treat infinite series like finite series.