If our population consists of $N$ individuals and $x_i$, $i=1,\ldots,N$, is the variable of interest, then the population mean and population variance are given by $ \bar{X}=\frac{1}{N}\sum_{i=1}^N x_i,\qquad \sigma^2=\frac{1}{N-1}\sum_{i=1}^N(\,x_i-\bar{X})^2. $ Suppose a simple random sample $x_1,\ldots,x_n$ of size $n$ is drawn from this population (i.e. every sample of size $n$ has equal probability of being drawn). Then the corresponding sample mean and sample variance is $ \bar{x}=\frac{1}{n}\sum_{i=1}^n x_i,\qquad \mathrm{var}(\bar{x})=\frac{1}{n-1}\sum_{i=1}^n (x_i-\bar{x})^2. $ It can be shown that $ \mathrm{var}(\bar{x})=\frac{N-n}{Nn}\sigma^2=\left(1-\frac{n}{N}\right)\frac{\sigma^2}{n}. $ Now, let $N\to\infty$ and see what happens.