convergence in distribution in $l_{2}$ Consider the space $l^{2}$. 
Let $X_{n}$ be a sequence of random vectors in $l^{2}$, such that $X_{n} \stackrel{d}{\to} X$. Let $Z_{n}$ be another sequence of random vectors in $l^{2}$, such that
$$
\mathbb{E}[\|X_{n} - Z{_n} \|_{2}^{2}] \to 0 \quad\text{as}\quad n\to\infty.
$$ 
The question: is it true that $Z_{n} \stackrel{d}{\to} X$, as $n\to \infty$?
 A: Yes: Write $Z_n=X_n+(Z_n-X_n)$; then $Z_n-X_n\to 0$ in probability. At this point you can appeal to Slutzky's theorem, formulated for a complete separable metric space such as $l_2$. As I don't have a ready reference for this extension, here's a proof.
Let $S$ be a complete separable metric space. Let $(X_n)$ and $(Y_n)$ be sequences of $S$-valued random variables defined on some probability space $(\Omega,\mathcal F,\Bbb P)$.
Suppose that $X_n\stackrel{d}{\to} X$ and $Y_n\stackrel{d}{\to} s$, where $s$ is some fixed element of $S$. Claim: $(X_n,Y_n)\stackrel{d}{\to}(X,s)$ (as random elements of $S\times S$). Once this is shown it will follow immediately from the continuous mapping theorem that (in the special case $S=l_2$) $X_n+Y_n\stackrel{d}{\to}X+s$. To prove the claim it suffices to show that $\lim_n\Bbb E[f(X_n,Y_n)]=\Bbb E[f(X,s)]$ for each bounded continuous function $f:S\times S\to\Bbb R$. Clearly $\lim_n\Bbb E[f(X_n,s)]=\Bbb E[f(X,s)]$ because $x\mapsto f(x,s)$ is bounded and continuous, and $X_n\stackrel{d}{\to} X$. It remains to show that
$$
\lim_n\Bbb E[f(X_n,Y_n)-f(X_n,s)]=0.
$$
Because of the assumed convergence in distribution, the sequence $\{(X_n,Y_n)\}_{n\ge 1}$ of pairs is tight. In particular, given $\eta>0$, there is a compact set $K\subset S$ such that $\Bbb P[X_n\in K,Y_n\in K]\ge 1-\eta$ for all $n$. Enlarging $K$ if necessary, we may suppose that $s\in K$. The restriction of $f$ to $K\times K$ is uniformly continuous, so given $\epsilon>0$ there exists $\delta>0$ such that if $(x,y)$ and $(x,y')$ are elements of $K$ with $d_S(y,y')<\delta$ then $|f(x,y)-f(x,y')|<\epsilon$.
Let $M$ be a bound for $f$. We then have
$$
\eqalign{
\Big|\Bbb E[f(X_n,Y_n)-f(X_n,s)]\Big|
&\le\Bbb E[|f(X_n,Y_n)-f(X_n,s)|]\cr
&\le 2M\Big\{\Bbb P[(X_n,Y_n)\notin K\times K]+\Bbb P[(X_n,Y_n)\in K\times K, d_S(Y_n,s)\ge\delta\Big\}\cr
&\phantom{bbbbbbbbbbbbb}+\epsilon\Bbb P[(X_n,Y_n)\in K\times K,d_S(Y_n,s)<\delta]\cr
&\le 2M\{\eta+\Bbb P[d_S(Y_n,s)\ge\delta]\}+\epsilon.\cr
}
$$
The upper bound above converges to $2M\eta+\epsilon$, which can be made as small as we please. The claim is proved.
