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.