Sum of Sequence 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.
 A: 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.
A: 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 


*

*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 

*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!)
