Let the random variables $X_1, X_2....X_n$ be independently Gaussian with mean, $\mathop{\mathbb{E}}[X_n]$ = $\mu sin(10n)$, and variance, $Var(X_n)$ = $\sigma^2(2+cos(10n))$. Does the sample mean $S_n=\frac{1}{n}\Sigma_{i=1}^nX_i$, converge to $\mu$ in probability. It's fact that, we use Chebyshev inequality to prove the convergence.
Chebyshev Inequality: $P(|S_n - \mathop{\mathbb{E}}[X_n]|\ge\epsilon)\le \frac{Var(S_n)}{\epsilon^2}$
As n->$\infty$, does RHS of inequality tends to 0 ?
obviously yes. $\mathbb{V}[S_n]=\frac{\mathbb{V}[X_n]}{n}$ thus in your RHS the numerator moves in $[\sigma; 3\sigma]$ while the denominator goes to $\infty$