Standard error of the sampling distribution of the mean I found an equation that says the standard error of the sampling distribution of the mean is:
$$\sigma_{\bar{X}} = \sigma \cdot \sqrt{\frac{1}{n}-\frac{1}{N}}$$
And when the population size is very large, the factor $1/N$ is approximately equal to zero; and the standard deviation formula reduces to:
$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$$
However, I did not get how he found the first equation, may someone explain it?
I found this equation here: http://stattrek.com/sampling/sampling-distribution.aspx
I could explain the second equation in that way:
\begin{equation}
\begin{split}
Var(\bar{X}) = Var\left(\frac{\sum_{i=1}^{n} X_i}{n}\right) \\
= \left(\frac{1}{n^2}\right)Var\left(\sum_{i=1}^{n} X_i\right) \\
= \left(\frac{1}{n^2}\right)\left(\sum_{i=1}^{n} Var(X_i)\right) \\
= \left(\frac{1}{n^2}\right)\left(\sum_{i=1}^{n}\sigma^2 \right) \\
= \left(\frac{1}{n^2}\right)\left(n\sigma^2\right) \\ = \frac{\sigma^2}{n}
\end{split}
\end{equation}
However I don't know from where appeared the first equation.
 A: This result relates specifically for finite population (of size $N$) and for sampling without replacement (for sample size $n$). It becomes clearer if we write
$$\sigma_{\bar{X}} = \sigma  \sqrt{\frac{1}{n}-\frac{1}{N}} = \frac {\sigma}{\sqrt n}\left(\sqrt {1-\frac nN}\right) = \frac {\sigma}{\sqrt n}\left(\sqrt {\frac {N-n}N}\right)$$
Comparing the correction factor with the formula provided in the CV post mentioned in the comment as a possible duplicate-maker of this one, Explanation of finite correction factor, there appears as $\left(\sqrt {\frac {N-n}{N-1}}\right)$. Why this difference in the denominator? 
@chl answer there mentions

You will notice that some authors use $N$ instead of $N-1$ in the
  denominator of the FPC; in fact, it depends on whether you work with
  the sample or population statistic: for the variance, it will be $N$
  instead of $N-1$ if you are interested in $S^2$ rather than
  $\sigma^2$.  

...which needs to be reconciled with @whuber's answer.
A: The quoted formula is not quite right.  Let's derive the correct one.
Since the population mean (or any other constant) may be subtracted from every value in a population $S$ without changing the variance of the population or of any sample thereof, we might as well assume the population mean is zero.  Letting the values in the population be $\{x_i\, \vert\, i\in S\}$, this implies
$$0 = \sum_{i\in S} x_i.$$
Squaring both sides maintains the equality, giving
$$0 = \sum_{i,j\in S}x_ix_j = \sum_{i\in S}x_i^2 + \sum_{i \ne j \in S} x_ix_j,$$
whence
$$\sum_{i\ne j \in S} x_ix_j = -\sum_{i\in S} x_i^2.$$
This key result will be employed later.
Let $S$ have $N$ elements. Because its mean is zero, its variance is the average squared value:
$$s^2 = \frac{1}{N}\sum_{i\in S}x_i^2.$$
(Please note that there can be no dispute about the denominator of $N$; in particular, it definitely is not $N-1$: this is a population variance, not an estimator.)
To find the variance of the sample distribution of the mean, consider all possible $n$-element samples.  Each corresponds to an $n$-subset $A\subset S$ and has mean
$$\frac{1}{n}\sum_{i\in A} x_i.$$
Since the mean of all the sample means equals the mean of $S$, which is zero, the variance of these $\binom{N}{n}$ sample means is the average of their squares:
$$s_n^2 = \frac{1}{\binom{N}{n}} \sum_{A\subset S}\left(\frac{1}{n}\sum_{i\in A}x_i\right)^2 = \frac{1}{n^2\binom{N}{n}} \sum_{A\subset S}\sum_{i,j\in A}x_ix_j \\= \frac{1}{n^2\binom{N}{n}} \sum_{A\subset S}\left(\sum_{i\in A}x_i^2 + \sum_{i\ne j\in A}x_ix_j\right) .$$
(Once again, $\binom{N}{n}$, not $\binom{N}{n}-1$, is the correct denominator: this is the variance of a collection of $\binom{N}{n}$ numbers, not an estimator of anything.)
Fix, for a moment, any particular index $i$.  The value $x_i$ will appear in $\binom{N-1}{n-1}$ samples, because each such sample supplements $x_i$ with $n-1$ more elements of $S$ out of the $N-1$ remaining elements (sampling is without replacement, remember).  Its contribution to the right hand side therefore equals $\binom{N-1}{n-1}x_i^2$.
Also fixing an index $j\ne i$, similar reasoning shows the product $x_ix_j$ appears in $\binom{N-2}{n-2}$ samples, thereby contributing $\binom{N-1}{n-1}x_ix_j$ to the right hand side.  Therefore, upon summing over all such $i$ and $j$ in $S$,
$$s_n^2 = \frac{1}{n^2\binom{N}{n}} \left(\binom{N-1}{n-1}\sum_{i\in S}x_i^2 + \binom{N-2}{n-2}\sum_{i\ne j\in S}x_ix_j\right).$$
Plug the first result into that last sum:
$$s_n^2 = \frac{1}{n^2\binom{N}{n}} \left(\binom{N-1}{n-1}\sum_{i\in S}x_i^2 + \binom{N-2}{n-2}\left(-\sum_{i\in S}x_i^2\right)\right).$$
It is now straightforward to relate this to the variance of $S$, because $\sum_{i\in S}x_i^2 = Ns^2$:
$$s_n^2 = \frac{1}{n^2\binom{N}{n}} \left(\binom{N-1}{n-1} - \binom{N-2}{n-2}\right)\left(Ns^2\right) = \frac{s^2}{n}\left(1 - \frac{n-1}{N-1}\right).$$
Thus the sampling variance for sampling with replacment, $\frac{s^2}{n}$, is multiplied by $1 - \frac{n-1}{N-1}$ to obtain the sampling variance for sampling without replacement, $s_n^2$.  Accordingly, the multiplicative adjustment for the sampling standard deviation is its square root, $\sqrt{1- \frac{n-1}{N-1}}$.  This differs from the quoted formula, which uses $\sqrt{1 - \frac{n}{N}}$.

Two simple checks can give us some comfort concerning the correctness of this result.  First, the sample variance of means of samples of size $n=1$, $s_1^2$, obviously equals the population variance $s^2$.  The correct formula states
$$s_1^2 = \frac{s^2}{1}\left(1 - \frac{1-1}{N-1}\right) = s^2,$$
as it should.  Unfortunately, the quoted formula asserts that $s_1^2 = s^2(\frac{1}{1} - \frac{1}{N})$ which obviously cannot be right.  Second, the sample variance of the means of samples of size $n=N$ is zero, because there is no variation, and indeed both formulas give $0$ in this case.
