In problem [1, p. 94 exercise 4.1] we will show that the periodogram is an inconsistent estimator by examining the estimator at $f=0$, or
 $\hat{P}_{PER}(0)=\frac{1}{N}\left(\sum\limits_{n=0}^{N-1}x[n]\right)^{2}.$ (1)

If $x[n]$ is real white Gaussian noise process with PSD
 $P_{xx}(f)=\sigma_{x}^{2}$ (2)

we are asked to find the mean an variance of $hat{P}_{PER}(0)$. We are asked if the variance converge to zero as $N \rightarrow \infty$. The hint provided within the exercise is to note that
 $\hat{P}_{PER}(0)= \sigma_{x}^{2} \left(\sum\limits_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}}\right)^{2}$ (3)

where the quantiiy inside the parenthesis is $\sim N(0,1)$ Solution: Because $Y = \sum_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}}$ is distributed according to a normal distribution $\sim N(0,1)$ the squared variable $Y^{2}=\frac{\hat{P}_{PER}(0)}{\sigma_{x}^{2}}$ is distributed according to a $\chi^{2}(1)$ with mean $E\left\{Y\right\}=1$ and variance $Var\left\{Y\right\}= 2$. From the previous relations we obtain the mean and the variance of the periodogram $\hat{P}_{PER}(0)$ as
 $E\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} E\left\{Y\right\} = \sigma_{x}^{2}$ $Var\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} Var\left\{Y\right\} = 2 \sigma_{x}^{2}.$

We see that while the mean converges to the true power spectral density, the variance does not converge to zero. Thus the estimator is inconsistent. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.