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A divergent series

By Tolaso in Uncategorized on May 20, 2017.

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

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

diverges.

Solution

We shall begin with a lemma.

Lemma: Let $x_1, \dots, x_n \in (0, 1)$ . It holds that

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

Proof: We are using induction on $n$ . For $n=1$ it is trivial. Suppose that it holds for $n=k$ . Then

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

Thus it holds for $n=k+1$ and the lemma is proved. Since $a_n>0$ and $\lim a_n =0$ I can find $1 = i_1 < i_2 < \cdots$ such that $\frac{a_{i_{r+1}}}{a_{i_r}} < \frac{1}{2}$ for all $r$ . But then

$\begin{align*} \sum_{n=1}^{i_k-1} \frac{a_n - a_{n+1}}{a_n} &= \sum_{r=1}^{k-1} \sum_{n=i_r}^{i_{r+1}-1}\left(1 - \frac{a_{n+1}}{a_n} \right) \\ &\geqslant \sum_{r=1}^{k-1} \left(1 - \frac{a_{i_{r+1}}}{a_{i_r}} \right) \\ &> \sum_{r=1}^{k-1} \frac{1}{2} \\ &> \frac{k-1}{2} \end{align*}$

where in the first inequality the lemma was used.

The exercise can also be found at mathematica.gr . It can also be found in Problems in Mathematical Analysis v1 W.J.Kaczor M.T.Nowak as exercise 3.2.43 page 80 .