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A divergent series

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

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



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 .


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