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You are quite close to solving the problem:

Using the representation $X_k=\sum_{j=1}^k Y_{kj}$ where $Y_{ij}\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$, you have $$\sum_{k=1}^n X_k= \sum_{k=1}^n\sum_{j=1}^k Y_{kj} = \sum_{u=1}^{n(n+1)/2} \xi_{u}$$where $\xi_u\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$ $(u=1,\ldots,n(n+1)/2)$. Therefore, if you normalise the above sum, you get $$\eqalign{\dfrac{\sum_{k=1}^n X_k-\mathbb{E}[\sum_{k=1}^n X_k]}{\text{var}(\sum_{k=1}^n X_k)^{1/2}} &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\mathbb{E}[\sum_{u=1}^{n(n+1)/2} \xi_{u}]}{\text{var}(\sum_{u=1}^{n(n+1)/2} \xi_{u})^{1/2}}\\ &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\frac{n(n+1)}{2} }{(n(n+1)/2)^{1/2}}\\ &=\sqrt{2}\,\dfrac{\sum_{k=1}^n X_k-\frac{n^2}{2}-\frac{n}{2}}{n(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{\frac{1}{n}\sum_{k=1}^n \left[X_k-\frac{n^2}{2}\right]-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{Z_n-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ }$$ which should help you conclude, along with a CLT on the above.

The theoretical references for a Central Limit Theorem for independent but not i.i.d. random variables are Liapounov's and Lindeberg's versions of the CLT. The former requires moments of order $2+\epsilon$ with $\epsilon>0$ and the latter for vanishing tail second moments. Both conditions apply for Poisson variates.

That the result holds (and hence that there is no mistake in the formulation) can be checked by a quick R experiment, as illustrated by the following that compares an histogram of 10³ $Z_n$'s with the $\text{N}(1/2,1/2)$ density:

You are quite close to solving the problem:

Using the representation $X_k=\sum_{j=1}^k Y_{kj}$ where $Y_{ij}\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$, you have $$\sum_{k=1}^n X_k= \sum_{k=1}^n\sum_{j=1}^k Y_{kj} = \sum_{u=1}^{n(n+1)/2} \xi_{u}$$where $\xi_u\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$ $(u=1,\ldots,n(n+1)/2)$. Therefore, if you normalise the above sum, you get $$\eqalign{\dfrac{\sum_{k=1}^n X_k-\mathbb{E}[\sum_{k=1}^n X_k]}{\text{var}(\sum_{k=1}^n X_k)^{1/2}} &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\mathbb{E}[\sum_{u=1}^{n(n+1)/2} \xi_{u}]}{\text{var}(\sum_{u=1}^{n(n+1)/2} \xi_{u})^{1/2}}\\ &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\frac{n(n+1)}{2} }{(n(n+1)/2)^{1/2}}\\ &=\sqrt{2}\,\dfrac{\sum_{k=1}^n X_k-\frac{n^2}{2}-\frac{n}{2}}{n(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{\frac{1}{n}\sum_{k=1}^n \left[X_k-\frac{n^2}{2}\right]-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{Z_n-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ }$$ which should help you conclude, along with a CLT on the above.

The theoretical references for a Central Limit Theorem for independent but not i.i.d. random variables are Liapounov's and Lindeberg's versions of the CLT. The former requires moments of order $2+\epsilon$ with $\epsilon>0$ and the latter for vanishing tail second moments. Both conditions apply for Poisson variates.

That the result holds (and hence that there is no mistake in the formulation) can be checked by a quick R experiment, as illustrated by the following that compares an histogram of 10³ $Z_n$'s with the $\text{N}(1/2,1/2)$ density:

You are quite close to solving the problem:

Using the representation $X_k=\sum_{j=1}^k Y_{kj}$ where $Y_{ij}\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$, you have $$\sum_{k=1}^n X_k= \sum_{k=1}^n\sum_{j=1}^k Y_{kj} = \sum_{u=1}^{n(n+1)/2} \xi_{u}$$where $\xi_u\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$ $(u=1,\ldots,n(n+1)/2)$. Therefore, if you normalise the above sum, you get $$\eqalign{\dfrac{\sum_{k=1}^n X_k-\mathbb{E}[\sum_{k=1}^n X_k]}{\text{var}(\sum_{k=1}^n X_k)^{1/2}} &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\mathbb{E}[\sum_{u=1}^{n(n+1)/2} \xi_{u}]}{\text{var}(\sum_{u=1}^{n(n+1)/2} \xi_{u})^{1/2}}\\ &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\frac{n(n+1)}{2} }{(n(n+1)/2)^{1/2}}\\ &=\sqrt{2}\,\dfrac{\sum_{k=1}^n X_k-\frac{n^2}{2}-\frac{n}{2}}{n(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{\frac{1}{n}\sum_{k=1}^n \left[X_k-\frac{n^2}{2}\right]-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{Z_n-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ }$$ which should help you conclude, along with a CLT on the above.

The theoretical references for a Central Limit Theorem for independent but not i.i.d. random variables are Liapounov's and Lindeberg's versions of the CLT. The former requires moments of order $2+\epsilon$ with $\epsilon>0$ and the latter for vanishing tail second moments.

That the result holds (and hence that there is no mistake in the formulation) can be checked by a quick R experiment, as illustrated by the following that compares an histogram of 10³ $Z_n$'s with the $\text{N}(1/2,1/2)$ density:

You are quite close to solving the problem:

Using the representation $X_k=\sum_{j=1}^k Y_{kj}$ where $Y_{ij}\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$, you have $$\sum_{k=1}^n X_k= \sum_{k=1}^n\sum_{j=1}^k Y_{kj} = \sum_{u=1}^{n(n+1)/2} \xi_{u}$$where $\xi_u\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$ $(u=1,\ldots,n(n+1)/2)$. Therefore, if you normalise the above sum, you get $$\eqalign{\dfrac{\sum_{k=1}^n X_k-\mathbb{E}[\sum_{k=1}^n X_k]}{\text{var}(\sum_{k=1}^n X_k)^{1/2}} &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\mathbb{E}[\sum_{u=1}^{n(n+1)/2} \xi_{u}]}{\text{var}(\sum_{u=1}^{n(n+1)/2} \xi_{u})^{1/2}}\\ &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\frac{n(n+1)}{2} }{(n(n+1)/2)^{1/2}}\\ &=\sqrt{2}\,\dfrac{\sum_{k=1}^n X_k-\frac{n^2}{2}-\frac{n}{2}}{n(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{\frac{1}{n}\sum_{k=1}^n \left[X_k-\frac{n^2}{2}\right]-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{Z_n-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ }$$ which should help you conclude, along with a CLT on the above.

The theoretical references for a Central Limit Theorem for independent but not i.i.d. random variables are Liapounov's and Lindeberg's versions of the CLT. The former requires moments of order $2+\epsilon$ with $\epsilon>0$ and the latter for vanishing tail second moments. Both conditions apply for Poisson variates.

That the result holds (and hence that there is no mistake in the formulation) can be checked by a quick R experiment, as illustrated by the following that compares an histogram of 10³ $Z_n$'s with the $\text{N}(1/2,1/2)$ density:

You are quite close to solving the problem:

Using the representation $X_k=\sum_{j=1}^k Y_{kj}$ where $Y_{ij}\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$, you have $$\sum_{k=1}^n X_k= \sum_{k=1}^n\sum_{j=1}^k Y_{kj} = \sum_{u=1}^{n(n+1)/2} \xi_{u}$$where $\xi_u\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$ $(u=1,\ldots,n(n+1)/2)$. Therefore, if you normalise the above sum, you get $$\eqalign{\dfrac{\sum_{k=1}^n X_k-\mathbb{E}[\sum_{k=1}^n X_k]}{\text{var}(\sum_{k=1}^n X_k)^{1/2}} &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\mathbb{E}[\sum_{u=1}^{n(n+1)/2} \xi_{u}]}{\text{var}(\sum_{u=1}^{n(n+1)/2} \xi_{u})^{1/2}}\\ &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\frac{n(n+1)}{2} }{(n(n+1)/2)^{1/2}}\\ &=\sqrt{2}\,\dfrac{\sum_{k=1}^n X_k-\frac{n^2}{2}-\frac{n}{2}}{n(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{\frac{1}{n}\sum_{k=1}^n \left[X_k-\frac{n^2}{2}\right]-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{Z_n-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ }$$ which should help you conclude. That the result holds can be checked by a quick R experiment, as illustrated by the following that compares an histogram of 10³ $Z_n$'s with the $\text{N}(1/2,1/2)$ density:  

You are quite close to solving the problem:

Using the representation $X_k=\sum_{j=1}^k Y_{kj}$ where $Y_{ij}\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$, you have $$\sum_{k=1}^n X_k= \sum_{k=1}^n\sum_{j=1}^k Y_{kj} = \sum_{u=1}^{n(n+1)/2} \xi_{u}$$where $\xi_u\stackrel{\text{iid}}{\sim}\mathcal{P}(1)$ $(u=1,\ldots,n(n+1)/2)$. Therefore, if you normalise the above sum, you get $$\eqalign{\dfrac{\sum_{k=1}^n X_k-\mathbb{E}[\sum_{k=1}^n X_k]}{\text{var}(\sum_{k=1}^n X_k)^{1/2}} &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\mathbb{E}[\sum_{u=1}^{n(n+1)/2} \xi_{u}]}{\text{var}(\sum_{u=1}^{n(n+1)/2} \xi_{u})^{1/2}}\\ &=\dfrac{\sum_{u=1}^{n(n+1)/2} \xi_{u}-\frac{n(n+1)}{2} }{(n(n+1)/2)^{1/2}}\\ &=\sqrt{2}\,\dfrac{\sum_{k=1}^n X_k-\frac{n^2}{2}-\frac{n}{2}}{n(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{\frac{1}{n}\sum_{k=1}^n \left[X_k-\frac{n^2}{2}\right]-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ &=\sqrt{2}\,\dfrac{Z_n-\frac{1}{2}}{(1+n^{-1})^{1/2}}\\ }$$ which should help you conclude, along with a CLT on the above.

The theoretical references for a Central Limit Theorem for independent but not i.i.d. random variables are Liapounov's and Lindeberg's versions of the CLT. The former requires moments of order $2+\epsilon$ with $\epsilon>0$ and the latter for vanishing tail second moments.

That the result holds (and hence that there is no mistake in the formulation) can be checked by a quick R experiment, as illustrated by the following that compares an histogram of 10³ $Z_n$'s with the $\text{N}(1/2,1/2)$ density: