## Incorrect Wikipedia Article?

Hey folks,

I'm currently studying sequences and the like in Calc 2, and I went to Wikipedia for another explaination about them. The example given in the article located here seems to be incorrect to me.

The example is this:

$\sum _{n=0}^{\infty }{2}^{-n} = 2$

I was thinking it's equal to zero though, since when n is really large, then the bottom gets really big so the whole fraction would head to zero.

Am I wrong or is the author wrong?
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 This is a series, and not a sequence.
 Yeah, that's what I meant to say ... but should that change the answer?

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## Incorrect Wikipedia Article?

Well, yes. A "series", a "sum of a series", a "sequence", and a "limit of a sequence" are all very different things.
 You're thinking of $\lim_{n \to \infty} 2^{-n}$, which is zero. However $\sum 2^{-n} = 1 + \frac{1}{2} + \frac{1}{4} + \cdots$ isn't zero.
 Yes, that answer is correct. $$1 + \frac{1}{2} + \frac{1}{4} + \cdots$$ does indeed equal 2. I probably solve simple series like this in a unique way, but I tend to imagine it in the number base 2 (binary). This would essentially be 1.11111111 repeating. This is like our 9.9999 repeating = 10, only that in binary is 2.

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 Quote by Weather Freak Am I wrong or is the author wrong?
Uhmmm, I am sorry to tell you this, but you are wrong, not the author...
This is a geometric series with the first term 1, and the common ratio r = 1 / 2.
So apply the formula to find the sum of the first n terms of a geometric series, we have:
$$S_n = a_1 \frac{1 - r ^ n}{1 - r}$$
Now r = 1 / 2. So |r| < 1, that means:
$$\lim_{n \rightarrow \infty} r ^ n = 0$$
Now let n increase without bound to get the sum:
$$\sum_{n = 0} ^ {\infty} 2 ^ {-n} = \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} a_1 \frac{1 - r ^ n}{1 - r} = \frac{a_1}{1 - r} = \frac{1}{1 - \frac{1}{2}} = 2$$.
Can you get this? :)
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@ KingNothing: Have you leant geometric series? We don't need to complicate the problem in binary, though. Just my \$0.02.
 Doh!! I get it now! Thanks folks... and I apologize for my confusion :(... it's all a little complicated when you first learn it.