Does $\sum^n_{k=1} (-1)^k$ converge for even $n$?

Question

This question was motivated by an misleading question of my own. If there already exists the same question (which I did not find), please mark this question as duplicate.

Take a look at the limit of this series: $$ \lim_{n\rightarrow\infty} \sum^{2n}_{k=1} (-1)^k. $$ Question: Does this limit exist? Does the series converge or diverge?

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Argumentation A

We can rewrite the sum as $$\lim_{n\rightarrow\infty} \sum^{2n}_{k=1} (-1)^k = \lim_{n\rightarrow\infty} \sum^{n}_{k=1} (-1 +1) = \lim_{n\rightarrow\infty} \sum^{n}_{k=1} 0 = 0.$$ Hence it converges.

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Argumentation B

We can rewrite the sum as $$\lim_{n\rightarrow\infty} \sum^{2n}_{k=1} (-1)^k = \sum^{\infty}_{k=1} (-1) = -1 + 1 - 1 + 1 - 1 + ....$$ Hence it diverges.

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Which argumentation is correct?

Answer 1

Argument A is correct.

In argument B the first equal sign is false. In general the notation $\sum_{k=1}^{\infty} a_k$ is shorthand for $\lim_{n\to\infty} \sum_{k=1}^n a_k$. So what you are claiming is that $\lim_{n\to\infty}\sum_{k=1}^n a_k = \lim_{n\to\infty}\sum_{k=1}^{2n}a_k$ which need not hold if both sides don't converge.

Answer 2

As $\forall n:S_{n}=0$, the sequence converges to $0$ (fairly quickly ).

The argumentation "B" doesn't account for the fact that you only consider the even numbers of terms.