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Let $c_{n} = \frac{1+(-1)^{n}}{2}$

$S_{n} = c_{1} + c_{2} + c_{3} + ... + c_{n}$

Prove that $\lim \frac{S_{n}}{n} = \frac{1}{2}$

These are my steps $\rightarrow \frac{S_{n}}{n} = \frac{n}{2n}(\frac{1+(-1)^{n}}{2}) = \frac{1+(-1)^{n}}{4}$

$1+(-1)^{n}$ is $2$ or $0$, so the lim of the sum is $\frac{2}{4} = \frac{1}{2}$

I dont know why, But I have the feeling that something wrong here. What do you think ?

Thanks.

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  • $\begingroup$ Your formula for $S_n$ is wrong. The sequence is $\{c_n\}=\{0,1,0,1,\dots\}$ so $S_{2n}=n=S_{2n+1}$. $\endgroup$
    – lulu
    Dec 13, 2015 at 12:00
  • $\begingroup$ I edited the post. $\endgroup$
    – NM2
    Dec 13, 2015 at 12:03
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    $\begingroup$ Still wrong. We have $S_n=\lfloor \frac n2 \rfloor\;\implies\;\frac {S_n}{n}=\frac 1n \lfloor \frac n2 \rfloor$ which does indeed go to $\frac 12$ (just look at the odd and even cases separately). $\endgroup$
    – lulu
    Dec 13, 2015 at 12:08
  • $\begingroup$ I understand your point. But this is in even case ,What about the odd case ? $\endgroup$
    – NM2
    Dec 13, 2015 at 12:12
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    $\begingroup$ My formula is true for all $n$. Separating the cases does make it easier to see the limit: $n=2k\;\implies\;\frac {S_n}{n}=\frac {k}{2k} = \frac 12$, while $n=2k+1\;\implies\;\frac {S_n}{n}=\frac {k}{2k+1} \to \frac 12$. Since both the odd terms and the even terms approach the same limit the entire sequence approaches that limit. $\endgroup$
    – lulu
    Dec 13, 2015 at 12:15

2 Answers 2

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To summarize the comments:

We are asked about the sequence $$c_n=\{0,1,0,1,\dots\}$$

It is easy to see that the partial sums satisfy $$S_{2n}=n=S_{2n+1}$$

To compute the limit (as $n\to \infty$) of $\frac {S_n}{n}$ it is convenient to distinguish the even indices from the odd.

If $n=2k$ is even we have $S_n=k$ from which we see that, in the even case, $$\frac {S_n}{n}=\frac 12$$.

If $n=2k+1$ is odd then we again have $S_n=k$ whence we see that we are trying to compute $$\lim_{n\to\infty} \frac {k}{2k+1}=\lim_{n\to\infty} \frac {1}{2+\frac 1k}=\frac 12$$

As both the odd and even terms of our sequence approach the same limit (namely $\frac 12$) the entire sequence approaches that limit and we are done.

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Morally, I think the lim(x-> infinity) (1+(-1)^n)/2=1 When n is 1, S(n)/n=0;2, S(n)/n=2;3, S(n)/n=0;4, S(n)/n=2....

So 2+0+2+0+2+0+...../1+1+1+1+1+1...

should be 1

Proof: Let k be the consecutive terms

Slope= (2+0)k/(1+1)k=2k/2k=1

Also we need not to proof that

  1. In geometric repesentation, this function passes through the point (0,0) in order to proof that numerator and denominator of 2+0+2+0+2+0+...../1+1+1+1+1+1... is equal.But in the sense of derivative we need not to prove it.(2k growing speed is always faster than a constant f(2k)'=2, f(c)'=0
  2. The sequence ends with the term odd. But in the sense of derivative we need not to prove it.(2k growing speed is always faster than a constant f(2k)'=2, f(2)'=0

(Sorry for a proof not so complete.Please help!)

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  • $\begingroup$ oh..sorry i see $\endgroup$ Dec 13, 2015 at 12:22

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