How to show that quadratic mean convergence implies expectation value? I am reading Larry Wasserman's All of Statistics and exercise 2 in chapter 6 asks for a proof that given sequence of random variables $ X_1, X_2, \dots $, show that $ X \xrightarrow{\text{QM}} b $ if and only if
$$
\begin{align}
& \lim_{n \rightarrow \infty} \mathbb{E}(X_n) = b & \text{and } & & \lim_{n \rightarrow \infty} \mathbb{V}(X_n) = 0.
\end{align}
$$
I'm getting stuck proving the forward direction. I started by expanding the definition of quadratic mean convergence as follows. By assumption, we have
$$
\lim_{n \rightarrow \infty} \mathbb{E}(X-b)^2 = 0.
$$
And then by linearity of expectation we have, 
$$
\lim_{n \rightarrow \infty} \mathbb{E}(X-b)^2 = \lim_{n \rightarrow \infty} \mathbb{E}(X_n^2) - 2b\ \mathbb{E}(X_n) + b^2 = 0.
$$
This is where I get stuck. It seems like we will somehow get that $ \mathbb{E}(X_n) $ has to equal $ b $ but I don't see how.
 A: By Jensen's Inequality, (alternatively, this follows from noting $\operatorname{Var}(X_n - b) \geq 0$
),$$\mathbb{E}(X_n - b)^2 \geq (\mathbb{E}|X_n - b|)^2$$
so taking the limit as $n\to\infty$ of both sides gives $0 \geq \limsup_{n\to\infty} \mathbb{E} |X_n - b|$, and we also clearly have $\liminf_{n\to\infty} \mathbb{E} |X_n - b| \geq 0$ since the argument is nonnegative. Then $\lim_{n\to\infty} \mathbb{E} |X_n - b| = 0$, so $\lim_{n\to\infty} \mathbb{E}(X_n) = b.$
For the second part, use the lemma posted from this stackexchange post. In particular, since $b$ is a constant, it has $0$ variance, so $\lim_{n\to\infty}\operatorname{Var}(X_n) = 0$.
A: The second part:
By the Var definition: 
$$Var(X) = E(X^2) - [E(X)]^2$$
$$Var(X_n-b) = E(|X_n-b|^2) - [E(X_n-b)]^2$$
$$\lim_{n \to \infty}⁡Var(X_n-b) = \lim_{n \to \infty}⁡E(|X_n-b|^2 ) - \lim_{n\to \infty}⁡{[E(X_n-b)]^2}$$
the Quadratic mean tell us that:
$$ \lim_{n\to \infty} ⁡E(|X_n-b|^2 ) = 0$$
and the first part of this problem defined:
$$\lim_{n \to \infty}⁡(E|X_n-b|) = 0,$$
then
$$\lim_{n \to \infty}⁡ Var(X_n-b) = 0.$$
