$p_n > 0$ and $p_{n+1} \ge p_n$. Prove $\sum \frac{ p_n-p_{n-1}}{p_np_{n-1}^a}$ is convergent where $a > 0$. $p_n > 0$ and $p_{n+1} \ge p_n$. Prove that
$$\sum_{n=1}^{\infty} \frac{p_n-p_{n-1}}{p_np_{n-1}^a}$$ is convergent where $a > 0$.
I could prove $\sum_{n=1}^{\infty} \frac{p_{n-1} - p_n}{p_n^{1+a}}$ is convergent since
$$\sum_{n=1}^{\infty} \frac{p_{n} - p_{n-1}}{p_n^{1+a}} \le \sum_{n=1}^{\infty}\int_{p_{n-1}}^{p_n} \frac{1}{x^{1+a}} \mathrm{d}x = \int_{p_0}^{\infty} \frac{1}{x^{1+a}} \mathrm{d}x $$
So I think this question may somehow be associated with the integral test.
 A: The case where $\{p_n\}$ is bounded and, hence, $p = \lim_{n \to \infty}p_n$ exists is relatively easy. We have 
$$\frac{p_n - p_{n-1}}{p_np_{n-1}^a} \leqslant \frac{1}{p_1p_0^a}(p_n - p_{n-1})$$
The series with terms on the RHS is telescoping. Thus, by the comparison test, 
$$\sum_{n=1}^\infty\frac{p_n - p_{n-1}}{p_np_{n-1}^a} \leqslant \frac{1}{p_1p_0^a}\sum_{n=1}^\infty(p_n - p_{n-1}) = \frac{1}{p_1p_0^a}(p - p_0)$$
On the other hand, if $\{p_n\}$ is unbounded, then $1/p_n \to 0$ as $n \to \infty$.
Since $a > 0$, there exists a positive integer $q$ such that $1/q < a$. Eventually, $p_{n-1} > 1$ and $p_{n-1}^{1/q} < p_{n-1}^a$.
Thus, there exists $m$ such that for all $n \geqslant m$ we have,
$$\frac{p_n - p_{n-1}}{p_np_{n-1}^a} < \frac{p_n - p_{n-1}}{p_np_{n-1}^{1/q}} = \frac{1 - \frac{p_{n-1}}{p_n}}{1 - \frac{p_{n-1}^{1/q}}{p_n^{1/q}}}\left(\frac{1}{p_{n-1}^{1/q}}- \frac{1}{p_n^{1/q}} \right)$$
If we can show that there exists a constant $C$ such that
$$\tag{*}\frac{1 - \frac{p_{n-1}}{p_n}}{1 - \frac{p_{n-1}^{1/q}}{p_n^{1/q}}} \leqslant C,$$
then we have convergence again by the comparison test, since in this case,
$$\sum_{n=m}^\infty\frac{p_n - p_{n-1}}{p_np_{n-1}^a} \leqslant C\sum_{n=m}^\infty\left(\frac{1}{p_{n-1}^{1/q}}- \frac{1}{p_n^{1/q}} \right) = C\left(\frac{1}{p_m^{1/q}} - \lim_{n \to \infty} \frac{1}{p_n^{1/q}} \right) = \frac{C}{p_m^{1/q}}$$
Proof of inequality (*)
Take $x = p_{n-1}^{1/q}/p_n^{1/q} = (p_{n-1}/p_n)^{1/q}$ and note that $0 < x \leqslant 1$.  
We have,
$$(1-x)q  \geqslant (1-x)\sum_{k=0}^{q-1}x^k = 1 - x^q,$$
which implies 
$$\frac{1 - \frac{p_{n-1}}{p_n}}{1 - \frac{p_{n-1}^{1/q}}{p_n^{1/q}}}= \frac{1- x^q}{1-x} \leqslant q$$
Therefore, the inequality (*) holds with $C = q$.
