The series diverges

Let $a_n$ be a decreasing sequence such that $a_n>0$ and $\lim a_n =0$. Prove that the series:

$$\sum_{n=1}^{\infty} \frac{a_n - a_{n+1}}{a_n}$$

diverges.

Solution

A quite handy lemma is the following:

Lemma: For $(0, 1) \ni x_i, \; i=1, \dots, n$ it holds that:

$$\sum_{i=1}^k (1-x_i) \geqslant 1 - \prod_{i=1}^k x_i$$

Proof:We are using induction. For $n=1$ the result is obvious. Suppose that it holds for $n=k$ then:

\begin{align*}\sum_{i=1}^{k+1} (1-x_i)  &= \left( \sum_{i=1}^{k} (1-x_i)\right) + (1-x_{k+1}) \\ &\geq \left(1 - \prod_{i=1}^k x_i\right) + (1-x_{k+1}) \\
&= 1 - \prod_{i=1}^{k+1} x_i + \left(1 - \prod_{i=1}^k x_i\right)\left(1-x_{k+1}\right) \\
&\geq  1 - \prod_{i=1}^{k+1} x_i
\end{align*}

and the proof is complete.

Since $a_n>0$ for each $n \in \mathbb{N}$ and $a_n \rightarrow 0$ we can find $1<i_1<i_2<\cdots$ such that

$$\frac{a_{i_{r+1}}}{a_{i_r}}< \frac{1}{2}, \;\; \forall r$$

Then this would imply

$$ \sum_{n=1}^{i_k-1} \frac{a_n - a_{n+1}}{a_n} = \sum_{r=1}^{k-1} \sum_{n=i_r}^{i_{r+1}-1}\left(1 - \frac{a_{n+1}}{a_n} \right) \geq  \sum_{r=1}^{k-1} \left(1 - \frac{a_{i_{r+1}}}{a_{i_r}} \right) >  \sum_{r=1}^{k-1} \frac{1}{2} > \frac{k-1}{2}$$

where in the first inequality the lemma was used. Thus, the series diverges.

The exercise can also be found at mathematica.gr