I have the following argument in my Time Series class notes:
Let ${u_t}$ be a mean zero covariance stationary process.
Define $\gamma(j) = \mathbb{E}u_tu_{t-j}$ and $Y_t = \mu +C(L)u_t$ where $C(L)=\sum_{j=0}^{\infty}c_jL^j$
The objective is that by using Chebyshev's inequality I can show that $P(|X - \mathbb{E}X| >\epsilon) \le Var(X)/\epsilon^2$ and therefore show that $n^{-1}\sum_{t=1}^n Y_t \rightarrow_p \mu$.
The answer to this question goes as follow:
$P(|X - \mathbb{E}X| >\epsilon) \le \frac{Var(n^{-1/2}\sum_{t=1}^{n}(Y_t - \mu))}{n\epsilon^2}$ by Chebyshev.
$\frac{Var(n^{-1/2}\sum_{t=1}^{n}(Y_t - \mu))}{n\epsilon^2} = \frac{2\pi fy(0)}{{n\epsilon^2}}$ where $fy(\lambda)$ denotes the spectral density of $Y$. If we denote the spectral density function of $u$ we have that
$fy(\lambda) = f_u(\lambda)\mid \sum_{j=0}^{\infty}c_je^{-i\lambda j}\mid^2$
and therefore $2\pi fy(0) = (\sum_{j=- \infty}^\infty \gamma(j)) (\sum_{j=0}^\infty c_j)^2$
My question is, where does the spectral density $2\pi fy(0)$ comes from? I would like to have an explanation as for why we can use the spectral density. Quite frankly, I don't really understand the implication of spectral density and its usage in this problem.
I guess the correction meant an inequality. Write that $$\operatorname{Var}\left(\sum_{t=1}^n(Y_t-\mu)\right)=C(L)\sum_{k=1}^n(n-k)\gamma(k),$$ which gives $$\operatorname{Var}\left(n^{-1/2}\sum_{t=1}^n(Y_t-\mu)\right)\leqslant C(L)\sum_{k=1}^n\gamma(k).$$ Now we can conclude if $\sum_k\gamma(k)$ is finite, which seems to be assumed.