<s>[INST] Checking my proof of some inequality I would like to receive some help about the next problem:
The Problem:
Prove that it is 
$$\frac{1}{n^{1 + \alpha}} < \frac{1}{\alpha} \left(\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right)$$, 
if we have that the next is valid: 
$$\frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} = \frac{-\alpha}{(n - \theta)^{\alpha + 1}} \;, 0 < \theta < 1$$,
where $\alpha \in \Bbb{R}$, $\alpha > 0$ and $n \in \Bbb{N} \setminus \{1\} $.
My solution:
$$ n - \theta < n \Rightarrow (n - \theta)^{\alpha + 1} < n^{\alpha + 1} \Rightarrow \frac{1}{(n - \theta)^{\alpha + 1}} > \frac{1}{n^{\alpha + 1}} \Rightarrow \frac{\alpha}{(n - \theta)^{\alpha + 1}} > \frac{\alpha}{n^{\alpha + 1}} \Rightarrow \frac{-\alpha}{(n - \theta)^{\alpha + 1}} < \frac{-\alpha}{n^{\alpha + 1}}.$$
$$ \frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} = \frac{-\alpha}{(n - \theta)^{\alpha + 1}} \Rightarrow 
\frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} < \frac{-\alpha}{n^{\alpha + 1}} \Rightarrow -\left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right) < \frac{-\alpha}{n^{\alpha + 1}} \Rightarrow 
\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} > \frac{\alpha}{n^{\alpha + 1}} \Rightarrow \frac{1}{\alpha} \left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right) > \frac{1}{n^{\alpha + 1}}.$$
Please, could you tell me is my solution correct and if not why?
Thank you, for your time and help!
 [/INST] We can't afford to have $n=0$, as $n$ appears in the denominator. Also, by removing $n=0$, we are sure that $n-\theta >0$ and hence you can raise positive power and preserve the inequality sign in the first line of the solution. 
Also, we can't have $n=1$ as $n-1$ appears in the denominator.
Other than that, that is if $n \ge 2$, the proof seems ok.
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