I would like to receive some help about the next problem:
In my textbook i have an example of use of the following theorem about the convergence of positive series:
Theorem:
For the members of the two positive series, $\sum_{n=0}^{\infty} a_n \; (1) $, $\sum_{n=0}^{\infty} b_n \; (2)$, let the next inequalities be correct: $a_n \le b_n$, ($n \ge n_0$, $n_0 \in \Bbb{N}$). In this case, the convergence of the series $(2)$ implicates the convergence of the series $(1)$ and the divergence of the series $(1)$ implicates the divergence of the series $(2)$.
Example:
Show that series $\sum_{n=0}^\infty \frac{1}{n^{1+\alpha}}$, $\alpha > 0$, converge.
Solution from the book:
Let us notice the series $\sum_{n=2}^\infty \left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right)$. Note that for the function $f(x) = \frac{1}{x^\alpha}$ we can use Mean value theorem on the segment $[n - 1, n]$. From there we get: $$ \frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} = - \frac{\alpha}{(n - \theta)^{\alpha + 1}} \;, 0 < \theta < 1.$$ From here we can get: $$ \frac{1}{n^{1 + \alpha}} < \frac{1}{\alpha} \left(\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right).$$ Partial sums $s_n$ of the series $\sum_{n=2}^\infty \frac{1}{\alpha} \left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right)$ are $s_n = \frac{1}{\alpha} \left( 1 - \frac{1}{n^\alpha}\right)$ and we can see that this series converge. So, series $\sum_{n=0}^\infty \frac{1}{n^{1 + \alpha}}$ also converge (according to already mentioned theorem - my addition).
What i don't understand:
I understand that in the beginning we got a series with positive members and that we need another if we want to use mentioned theorem as a criteria for convergence. I understand how with the help of the Mean value theorem we can prove: $$ \frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} = - \frac{\alpha}{(n - \theta)^{\alpha + 1}} \;, 0 < \theta < 1,$$ and how we got: $$ \frac{1}{n^{1 + \alpha}} < \frac{1}{\alpha} \left(\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right).$$
Now, because the second series converge to $1$, i don't understand why that series wasn't used in the conclusion of the example instead of the third series with $\frac{1}{\alpha}$? We have that following is correct: $$ \frac{1}{n^{1 + \alpha}} < \frac{1}{\alpha} \left(\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right) < \left(\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right)$$ and it gives us fulfilled condition for using the theorem.
Please, could you tell me where i make a mistake?
Thank you, for your time and help!
EDIT (4.3.2018. :))
I think i overthought this problem. In the given solution, they didn't try to use the series $\sum_{n=2}^\infty \left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right)$ to compare it with the given series. They just used that series and the effect of the Mean value theorem on the function $f(x) = \frac{1}{x^\alpha}$, to construct the series $\sum_{n=2}^\infty \frac{1}{\alpha} \left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right)$ and compare it with the given series.
I checked if $\frac{1}{\alpha}\left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha}\right) < \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha}$ is correct for $\alpha = \frac{1}{2}$ and it isn't.
In the book there isn't a proof for $\frac{1}{n^{1 + \alpha}} < \frac{1}{\alpha} \left(\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right)$, but i think that this can be a proof:
$$ n - \theta < n \Rightarrow (n - \theta)^{\alpha + 1} < n^{\alpha + 1} \Rightarrow \frac{1}{(n - \theta)^{\alpha + 1}} > \frac{1}{n^{\alpha + 1}} \Rightarrow \frac{\alpha}{(n - \theta)^{\alpha + 1}} > \frac{\alpha}{n^{\alpha + 1}} \Rightarrow \frac{-\alpha}{(n - \theta)^{\alpha + 1}} < \frac{-\alpha}{n^{\alpha + 1}}.$$
$$ \frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} = \frac{-\alpha}{(n - \theta)^{\alpha + 1}} \Rightarrow \frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} < \frac{-\alpha}{n^{\alpha + 1}} \Rightarrow $$ $$ -\left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right) < \frac{-\alpha}{n^{\alpha + 1}} \Rightarrow \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} > \frac{\alpha}{n^{\alpha + 1}} \Rightarrow$$ $$ \frac{1}{\alpha} \left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right) > \frac{1}{n^{\alpha + 1}}.$$
Please, could you tell me if this is correct solution and if not why?
Thank you for your time and help!