TITLE: $CLT$ and $LLN$ give different results
QUESTION [10 upvotes]: I tried to solve a problem two different ways and I got different results.
Let $( X_i )_{i \in \mathbb{N}}$ be a series of independent, identically distributed random variables, with $\mathbb{E}[X_i] = 1$ and $\mathbb{V}[X_i] = 1$
Determine
$$
\lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right)
$$
Here are the two approaches that I tried.
Central limit theorem
\begin{align*}
& \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) \\
= {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - 1) \leq \sqrt{n} - \frac{n}{\sqrt{n}}\right) \\
= {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - 1) \leq 0\right) \\
= {} & \Phi_{0,1}(0) = \frac{1}{2}
\end{align*}
Law of large numbers
\begin{align*}
& \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) \\
= {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n X_i \leq 1 \right) \\
\geq {} & \lim_{n \to \infty} \mathbb{P}\left(\frac{1}{n} \sum_{i=1}^n X_i = 1\right) \\
= {} & 1
\end{align*}
according to the strong law of large numbers. This then means that
$$
\lim_{n \to \infty} \mathbb{P}\left(\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \leq \sqrt{n}\right) = 1
$$
What am I doing wrong here? My understanding is that the CLT solution is correct, but I don't see what I did wrong with applying the law of large numbers either.

REPLY [16 votes]: What the (strong) law of large numbers says: $$ \require{cancel} \begin{align} \text{right: } & \Pr\left( \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n X_i  = 1\right) = 1. \\  \\ \text{wrong: } & \xcancel{\lim_{n\to\infty} \Pr\left( \frac 1 n \sum_{i=1}^n X_i  = 1\right) = 1.} \end{align} $$