I am currently doing some exercises for an exam in probability theory and am stuck with the following exercise:
Let $(X_n)$ be random variables which are binomialdistributed with parameter $n,p$ for $p\in (0,1)$. Show that \begin{align} \frac{X_n(1+\frac{1}{\sqrt{n}})-np}{\sqrt{n}}\Rightarrow \mathcal{N}(p,p(1-p)) \end{align} where $\Rightarrow$ denotes convergence in distribution. My try was to insert $S_n-S_n$ and split the sum which with the central limit theorem implies that one summand converges in distribution to $\mathcal{N}(0,p(1-p))$. EDIT: This is false.
It remains to show that $\frac{X_n(1+\frac{1}{\sqrt{n}})-S_n}{\sqrt{n}}$ converges to $p$ in probability and this is were I am stuck. Any help is appreciated. Thanks in advance
You have $$\frac{X_n(1+\frac{1}{\sqrt{n}})-np}{\sqrt{n}} = \sigma\frac{X_n - np}{\sigma \sqrt{n}} + \frac{X_n}{n}$$ for $\sigma = \sqrt{p(1-p)}$
Additionally we have $$X_n \sim \sum_{k=1}^n Y_k$$ for $Y_k \sim Ber(p)$ So we get $$\frac{X_n - np}{\sigma \sqrt{n}} \sim \frac{\sum_{k=1}^n Y_k - np}{\sigma \sqrt{n}} \Rightarrow \mathcal{N}(0,1)$$ by the CLT and it's known that $$\frac{X_n}{n} \to p$$
So we get in total: $$\frac{X_n(1+\frac{1}{\sqrt{n}})-np}{\sqrt{n}} \Rightarrow \sigma\mathcal{N}(0,1) + p = \mathcal{N}(p,\sigma^2)$$