Convergence in Distribution implies Convergence in Expectation? I need some help the following problem: 

Let $X_1,X_2,...$ be a sequence of random variables such that $X_n$ converges to $X$ in distribution, where $X \sim \text{Poisson}(\lambda)$. Let the function $f(x)=I_{(1,10)}(x)$ be the indicator function. Do we have $$E[f(X_n)]\to E[f(X)]?$$Hint: $X_n=X+1/n.$

What I tried is the following: 
$$E[f(X_n)]=\sum_{n=0}^{\infty}I_{(1,10)}(x)e^{-\lambda} {{\lambda}^{x+1/n} \over {(x+1/n)!}}$$ and $$E[f(X)]=\sum_{n=0}^{\infty}I_{(1,10)}(x)e^{-\lambda} {{\lambda}^{x} \over {x!}}.$$
Then evaluate their difference: 
$E[f(X_n)]-E[f(X)]=\sum_{n=0}^\infty I_{(1,10)}(x)e^{-\lambda}\big( {{\lambda}^{x+1/n} \over {(x+1/n)!}}-{{\lambda}^{x} \over {x!}}\big)$. 
But I don't feel this will give something helpful if I move on. Am I on the right track? Or how to solve this problem in some other way? 
 A: Why are you considering the sum of your rv's? Also, the statement of the problem does not say that the r.v's in the sequence follow the Poisson, only that the sequence converges to a Poisson. But I will assume that the $X_n$'s are discrete.
Your $f$ is bounded. If it was also continuous, this result would be part of the "Portmanteau Theorem" (or Lemma). Now that $f$ is not continuous, we make use of the fact that $E(I_{A}(x)) = P(x \in A)$
So the question can be re-written as
$$P(X_n\in (1,10)) \rightarrow ? \; P(X\in (1,10))$$
Since you have used parentheses for $(1,10)$, I assume that $(1,10) = \{1,2,...,10\}$ (although it doesn't make any difference). So we require whether
$$\sum_{1}^{10}P(X_n=i) \rightarrow? \sum_{1}^{10}P(X=i)$$
But you have already assumed that $X_n$ converges in distribution to $X$. Convergence in distribution for discrete rv's is equivalent (see for example this post)  to the fact that $$P(X_n=i) \rightarrow P(X=i)\qquad \forall i$$
Since by convergence in distribution we have essentially assumed that for each point in the support, the pmf converges to a corresponding constant (the limiting probability for each $i$), the sum of some of these points in the support (LHS of the question) will converge to the sum of these constants (RHS of the question). QED.
