Suppose that $X_n$ are iid r.v.'s with mean $1$
and $a_n$ are real numbers with $\lim_{n} a_n=1$.
Show that $ n^{-1}\sum_{j=1}^{n}a_jX_j \rightarrow 1$ almost surely.
There is no assumption about $X_n$ except for there means, so I tried to use strong law of large numbers, but I didn't know what to do with those $a_n$s.
I don't know any different version of SLLN.
By the strong law of large numbers, we have
$$\frac{1}{n} \sum_{j=1}^n X_j \to \mathbb{E}(X_1)=1 \quad \text{a.s.}$$
and therefore it suffices to show that
$$\frac{1}{n} \sum_{j=1}^n (a_j-1) X_j \to 0 \qquad \text{a.s.}$$ For fixed $\epsilon>0$ we can choose $N \in \mathbb{N}$ such that $|a_j-1| \leq \epsilon$ for all $j \geq N$. Clearly,
$$\frac{1}{n} \sum_{i=1}^n (a_j-1) X_j = \frac{1}{n} \sum_{j=1}^N (a_j-1) X_j + \frac{1}{n} \sum_{j=N+1}^n (a_j-1) X_j.$$
The first term on the right-hand side converges to $0$ as $n \to \infty$. Prove that the (modulus of the) second term is bounded by
$$\epsilon \frac{1}{n} \sum_{j=N+1}^n |X_j| \leq \epsilon \frac{1}{n} \sum_{j=1}^n |X_j|.$$
Use the SLLN to show that this expression is bounded by $\epsilon \mathbb{E}(|X_1|)$ as $n$ tends to $\infty$. Conclude.