# Strange Sequence

## Homework Statement

Find sequence $$(a_n)$$ s.t. $$\lim_{N\rightarrow \infty} \sum_{n=1}^{2N} a_n$$ and $$\lim_{N\rightarrow \infty} \sum_{n=1}^{2N+1} a_n$$ both converge but $$\sum_{n=1}^{\infty} a_n$$ diverges.

I have no idea where to start to be honest. I'm confused at how this is possible. Isn't it always just summed to infinity regardless of whether it's 2(x) or 2(x+1) where x-> infinity?

## Answers and Replies

$$a_n=(-1)^n$$

Can you explain how that works? I'd rather know what the problems means/is asking instead of really knowing the answer. I was guessing it had to do with alternating series, but I don't know what differentiates 2N, 2N + 1, and N.

Or does it mean that:
for the first sequence
(-1)^n = -1 + 1 + -1 + ... + (-1)^(2N) = (-1)^(2N) -> 1 as n approaches infinity
and same logic for the rest?
That would make sense I guess.

Edit: actually since there's an even number of terms for the first sequence and it goes -1,1, etc., then it should converge to 0. I would assume this is the correct logic to follow, but how is shown rigorously? Clearly, there must be something wrong in saying there is an even # of infinite terms.

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$$\sum_{n=1}^{N}a_n=(-1)^N$$
so the series doesn't converge. On the other hand
$$\sum_{n=1}^{2N}a_n=0\qquad\sum_{n=1}^{2N+1}a_n=-1$$
so both limits exist. Sorry I left it w/o explaination, I though it was this kind of problems where the answer is kinda tricky, but easy to verify.

Yeah, after a bit longer, I recognized the logic. Thanks a bunch! Is there a way to rigorously explain that $$\lim_{N\rightarrow \infty} \sum_{n=1}^{2N} a_n=0$$?

Since each member of this sequence is zero, the limit is zero as well.

I'm not really sure how you define member since in my belief, each individual term is -1 or 1. However, I'm assuming you mean member is not the same as term.
Is it valid to say $$\lim_{N\rightarrow \infty} \sum_{n=1}^{2N} a_n= (-1) + 1 + (-1) + 1 + ... = (-1 + 1) + (-1 + 1) + ... = \sum_{n=1}^{\infty} (-1+1) = \sum_{n=1}^{\infty} 0 = 0$$?

$$\sum_{n=1}^{2N}a_n=0$$ for every $$N$$, and so,
$$\lim_{N\rightarrow\infty}\sum_{n=1}^{2N}a_n=\lim_{N\rightarrow\infty}0=0$$

Oh. Makes sense. Thanks.