Convergence in $L^1$ of $X_n(z):=n^{\alpha}1_{[\frac{1}{n+1}, \frac{1}{n}]}(z)$

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Let $X_n(z):=n^{\alpha}1_{[\frac{1}{n+1}, \frac{1}{n}]}(z)$ with $\alpha>0$ on $[0,1]$ be a series.

I have two questions about this.
For which $\alpha$ does $X_n$ converge in $L^1$ to 0 with $n \rightarrow \infty$?
For which $\alpha$ can I find an integrable dominating convergence?

I know that $X_n$ converges pointwise to 0 for $n \rightarrow \infty$.

But I don'r really know how to go about it. A nudge into the right direction would help a lot.

My first thought was to show the second one first to get a Lebesgue-integrable dominating function. The Lebesgue theorem would provide the first statement, right?

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If $(X_n)_{n\geqslant 1}$ converges in $\mathbb L^1$ to a function $X$, then necessarily $X=0$ (because we extract a subsequence $(X_{n_k})_{k\geqslant 1}$ which converges almost everywhere and identify the limits).

The computation of $\lVert X_n\rVert_1$ gives $n^{\alpha-1}/(n+1)$.

If there is $Y$ which dominates each $X_n$ and $Y$ is integrable, then the function $x\mapsto \sup_nX_n(x)$ should be integrable. The supports of $X_n$'s are almost disjoint (up to sets of measure $0$) hence the task is to check whether $\sum_{n\geqslant 1}n^{\alpha}\chi_{[1/(n+1),1/n)}$ is integrable.