Show that ${X_n}$ is i.i.d and $E(X_1) = 1$ and $c_n$ is bounded real numbers
then $\frac{c_1 + ... + c_n}{n} \rightarrow 1$ if and only if $\frac{c_1X_1 + ... + c_nX_n}{n} \rightarrow 1$.
In the K.L. Chung A course in probability theory(3rd edition), there is hint s.t. truncate Xn at n and proceed as in Theorem 5.4.2(the Strong law of Large numbers).
But by separating integrals with the area of $j-1<|c_nX_n|<j$, since c_n can be 0 and cn is not same, applying the process of the proof of SLLN is not easy. Is there any another way to prove this?
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Thank you for your answer. I solved it using another way.
As in the K.L.chung, define Yn = XnI(|xn| <= n), not Yn = anXn* I(|xn| <= n) Then, the infinite sum of Var(anYn/n) converges similar to the proof of SLLN.
It is a consequence of the (usual) law of large numbers, that is, $S_n/n\to\mathbb E\left[X_1\right]$, where $S_n=\sum_{i=1}^nX_i$. Indeed, \begin{align} \frac 1n\sum_{i=1}^nc_iX_i &=\frac{c_n}nS_n+\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\frac{S_j}j\\ &=\frac{c_n}nS_n+\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\left(\frac{S_j}j-1\right)+\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\\ &=\frac{c_n}nS_n+\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\left(\frac{S_j}j-1\right)+\frac 1n\sum_{j=1}^{n-1}j c_j -\frac 1n\sum_{j=2}^n\left(j-1\right)c_j \\ &=\color{blue}{\frac{c_n}nS_n-\frac{n-1}nc_n } +\frac 1n\sum_{j=1}^nc_j +\frac 1n\sum_{j=1}^{n-1}j\left(c_j-c_{j+1}\right)\left(\frac{S_j}j-1\right). \end{align}
Now, the conclusion follows if we manage to show that