Weighted sum of large numbers From the law of large numbers, if $X_1,X_2,...X_N$ are i.i.d random variable, then we have
$$\lim_{N \rightarrow \infty} \frac{1}{N} (\Sigma_1^N X_i)=\mu$$
where $\mu$ is the mean of $X_i$.
What I want to ask is if there is a infinite sequence $a_i$.
Can we express the value of $\lim_{N \rightarrow \infty} \frac{1}{N} (\Sigma_1^N a_iX_i)$ with $a_i$?
 A: The result regarding the expectation is actually quite simple:
$$\lim_{N\to\infty}E[\frac1N\sum_1^Na_nX_n]=\lim_{N\to\infty}\frac1N\sum_1^Na_n\mu=\mu\bar a,$$
where $\bar a=\lim_{N\to\infty}\frac1N\sum_1^Na_n$, assuming this limit exists. However, it doesn't come with as much force as the real law of large numbers, because it doesn't say anything about the variance. For that, we can use the following extension to LLN known as Kolmogorov's strong law and mentioned at the Wikipedia article:

If $X_n$ are independent (not necessarily i.i.d.), then $\bar X_n-E[\bar X_n]\overset{a.s.}{\to}0$, provided that $\sum_{k=1}^\infty\frac1{k^2}\operatorname{Var}[X_k]<\infty$.

From this, assuming that $\operatorname{Var}[X_i]=\sigma^2$ is finite, if we set $X_n\mapsto a_nX_n$ in this law we get
$$P\left(\lim_N\frac1N\sum_{n=1}^Na_n(X_n-\mu)=0\right)=1$$
$$P\left(\lim_N\frac1N\sum_{n=1}^Na_nX_n-\mu\bar a=0\right)=1$$
$$P\left(\lim_N\frac1N\sum_{n=1}^Na_nX_n=\mu\bar a\right)=1$$
so $\lim_N\frac1N\sum_{n=1}^Na_nX_n\overset{a.s.}{\to}\mu\bar a$, provided that $\sum_{k=1}^\infty\frac{a_k^2}{k^2}<\infty$.

Regarding the distribution of the infinite sum $\sum_na_nX_n$, it satisfies no law of large numbers analogue, because there is no smoothing factor. The variance only gets worse after you add each term. So it won't converge almost surely to anything unless the distribution of $X_i$ is trivial (concentrated at one point). However, it will be true that the expectation is $E[\sum_na_nX_n]=\mu\sum_na_n$ and the variance is $\operatorname{Var}[\sum_na_nX_n]=\sigma^2\sum_na_n^2$, assuming these are finite.
