Prove $\sum_{n=1}^\infty(\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})$ convergent

Question

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!

Answer 1

HINT:

Since $a_n$ is positive and increasing , then $\frac1{a_n}$ is positive and decreasing. Moreover, we have

$$\sum_{n=1}^{2N}\frac{(-1)^{n-1}}{a_n}=\sum_{n=1}^N \left(\frac1{a_{2n-1}}-\frac1{a_{2n}}\right)$$

Now use what you know about $\lim_{n\to \infty}a_n$. Can you finish now?

Answer 2

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.

Answer 3

Hint: Let

$$b_n=\frac1{a_{n-1}}-\frac1{a_{n}}$$

Each $b_n$ is non-negative, since $a_n$ is increasing.

Then your series is: $$\sum_{n=1}^\infty b_{2n}$$ But:

$$\sum_{n=2}^{2N} b_n\geq \sum_{n=1}^{N} b_{2n}\geq 0$$

And $$\sum_{n=2}^{2N} b_n=\frac1{a_{1}}-\frac1{a_{2N}}.$$