Consider $(x_i)_{i=1}^n$ a sequence of i.i.d random variables with mean $\mu$ and variance $\sigma ^ 2$. I want to prove that:
$$\frac{2}{n(n+1)}\sum_{i=1}^{n}ix_i \overset{p}{\to} \mu$$
I've tried this using the Law of Large Numbers, however I am not sure about how to deal with the term $i$ inside the sum. Any suggestions on how to prove this?
Let $S_n=\frac{2}{n(n+1)}\sum_{i=1}^niX_i$, then $$ \mathbb{E}[S_n]=\frac{2}{n(n+1)}\sum_{i=1}^ni\mathbb{E}[X_i]=\frac{2\mu}{n(n+1)}\sum_{i=1}^ni=\mu $$ using the linearity of expectation and the identity $\sum_{i=1}^ni=\frac{n(n+1)}{2}$.
Similarly, since the $X_i$ are i.i.d. we have $$ \mathrm{var}(S_n)=\frac{4}{n^2(n+1)^2}\sum_{i=1}^ni^2\mathrm{var}(X_i)=\frac{4\sigma^2}{n^2(n+1)^2}\sum_{i=1}^ni^2=\frac{2(2n+1)\sigma^2}{3n(n+1)} $$ using the identity $\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$.
Now recall that for any $\varepsilon>0$, Chebyshev's inequality states that $$ \mathbb{P}(|S_n-\mathbb{E}[S_n]|>\varepsilon)\leq \frac{\mathrm{var}(S_n)}{\varepsilon^2}$$ Plugging in the above results into Chebyshev's inequality yields $$ \mathbb{P}(|S_n-\mathbb{E}[S_n]|>\varepsilon)\leq \frac{2(2n+1)\sigma^2}{3\varepsilon^2n(n+1)}\to 0 $$ as $n\to\infty$ for any fixed $\varepsilon>0$, so $S_n\to \mu$ in probability.