Prove $\sum_{n=1}^\infty(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})$ convergent Let $(a_n)_{n=1}^\infty$  Let be a positive, increasing, and unbounded sequence. Prove that the series:
$$\sum_{n=1}^\infty\left(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}}\right)$$
convergent.

We know that since $a_n$ is increasing and unbounded, than $\lim_{n \to \infty}a_n=\infty$, so I want to apply that to say that, $\sum_{n=1}^\infty(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})$ is decreasing and its limit will be $0$.
My problem is that every time I get confused while it says $a_{2n}$ or $a_{2n-1}$, it is less intuitive for me than just "normal" $a_n$...
Appreciate your help!
Thanks a lot!

The fact that $(a_n)$ is unbounded is actually not needed, only that the sequence is positive and increasing.
First note that all terms $b_n = \frac{1}{a_{2n-1}}-\frac{1}{a_{2n}}$ are non-negative. Then verify that
$$
 \sum_{n=1}^N b_n = \frac{1}{a_1} + \underbrace{\left(-\frac{1}{a_2}+\frac{1}{a_3}\right)}_{\le 0}+  \cdots
+\underbrace{\left(-\frac{1}{a_{2N-2}}+\frac{1}{a_{2N-1}}\right)}_{\le 0}
 -\frac{1}{a_{2N}} \le \frac{1}{a_1} \, .
$$
So the partial sums of $\sum_{n=1}^\infty b_n$ are increasing and bounded above.  It follows that the series is convergent.
