I am currently trying to solve a question and I think I know how to proceed but am uncertain. The problem reads as follows:
Let $X_1, X_2, \dots, X_n$ be i.i.d. random variables with finite second moment, i.e. $\mathbb{E}[X_1^2] = a^2$. Assume that $\mathbb{E}[X_1] = m$. Let: $$T_n :=\frac{2}{n(n+1)}\sum_{i=1}^{n} iX_i$$Then show that $T_n \rightarrow m$ in probability.
My way of showing that $T_n \rightarrow m$ in probability is to note that $$\sum_{i=1}^{n}\frac{X_i}{n} \rightarrow m$$ in probability by the Weak Law of Large Numbers. Then is what is left to show that: $$\sum_{i=1}^{n}\frac{2i}{n+1} \rightarrow 1$$in probability? I'm confused by the second part because it doesn't really make sense, but I'm not sure how else to tackle this question.
WLLN says that if $Z_1,\dots,Z_n$ are iid with finite mean $\mu < \infty$, then $\dfrac{1}{n} \displaystyle \sum_{i=1}^n Z_i$ converges in probability to $\mu$.
I find this hard to apply in your case, since the "summation analogue" is $\dfrac{1}{n} \displaystyle\sum_{i=1}^n \dfrac{2 i X_i}{n+1}$, and here the random variables $2iX_i /(n+1)$ aren't iid since they no longer have the same mean.
Instead, you can use Chebyshev. Let $\epsilon>0$ be given. Note that $$ E\left[ T_n \right] = E \left[ \frac{2}{n(n+1)}\sum_{i=1}^{n} iX_i \right] = \dfrac{2}{n(n+1)} \sum_{i=1}^n i E[X_i] = \dfrac{2 m}{n(n+1)} \sum_{i=1}^n i = m. $$
So by Chebyshev $ P(|T_n - m| \ge \epsilon) \le \dfrac{\text{Var } T_n}{\epsilon^2}. $ If we can show $\text{Var } T_n \xrightarrow{n\to\infty}0$, then we're finished. \begin{align*} \text{Var } T_n &= \dfrac{4}{(n(n+1))^2} \text{Var } \left( \sum_{i=1}^n i X_i \right)\\ &\stackrel{ind.}{=}\dfrac{4}{(n(n+1))^2} \sum_{i=1}^n \text{Var } \left(i X_i \right) \\ &=\dfrac{4a^2}{(n(n+1))^2} \sum_{i=1}^n i^2 \\ &= \dfrac{4a^2}{(n(n+1))^2} \left( \dfrac{n}{6} + \dfrac{n^2}{6} + \dfrac{n^3}{3} \right) \end{align*} which indeed tends to zero as $n \to \infty$.