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Let $(a_n)$ be a sequence of positive real numbers, Such that $(a_1+a_3+a_5+\dots +a_{2n-1})(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_{2n-1}})\geq(2n-1)^2$.Then which of the following is/are true?

1.$(a_n)$ is a Cauchy Sequence.

2.$(a_n)$ can be a cauchy sequence.

3.$(a_n)$ is a monotonic sequence.

4.$(a_n)$ has a convergent subsequence.

I tried to construct some counters to discard the options but eventually most of them were not satisfying the given condition. For instance in order to discard (3) we can have $(a_n)=(2,3,2,3,\dots) $(but this is not correct!). A sequence of real numbers is convergent iff its cauchy and to have a convergent subsequence $(a_n)$ should be bounded too.How can I use the given condition to get some conclusion out? Any help? Thanks.

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    $\begingroup$ Here's a hint: consider the sequence $a_n=\{3,1,5,1,7,1,9,1,...\}$. $a_n=2n+1$ if $n$ is odd and $1$ if $n$ is even. $\endgroup$
    – QC_QAOA
    Jun 27, 2022 at 6:09
  • $\begingroup$ Thanks @QC_QAOA it discarded (3) and (1) both but (2) is the only correct answer. Any idea for (4)? $\endgroup$ Jun 27, 2022 at 8:56

1 Answer 1

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Too long for a comment:

To show $2)$ is true, consider $a_n=\frac{1}{n^3}$. Then according to Mathematica

$$\lim_{n\to\infty}\frac{\left[\sum_{i=1}^{n}a_{2i-1}\right]\left[\sum_{i=1}^{2n-1}\frac{1}{a_i}\right]}{(2n-1)^2}=\infty$$

and holds true for all $n$.

To show $4)$ is false, consider the sequence

$$a_n=\begin{cases} n & \text{if } n\text{ is even}\\ n^2 & \text{if } n\text{ is odd} \end{cases}$$

Then again, according to Mathematica

$$\lim_{n\to\infty}\frac{\left[\sum_{i=1}^{n}a_{2i-1}\right]\left[\sum_{i=1}^{2n-1}\frac{1}{a_i}\right]}{(2n-1)^2}=\infty$$

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  • $\begingroup$ For $(a_n)=\frac{1}{n}$, let $n=3$ then we have $(a_1+a_3)(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3})\geq25\implies$ $8\geq25$(not satisfying the given condition).Interpreting something wrong? Can you get those links of Mathematica into your answer, please. $\endgroup$ Jun 27, 2022 at 19:35
  • $\begingroup$ Sorry, I was only checking the limiting behavior of the sequence. To make sure it works for all $n$, set $a_n=\frac{1}{n^2}$ instead. $\endgroup$
    – QC_QAOA
    Jun 27, 2022 at 19:44
  • $\begingroup$ Please don't be sorry, I am glad you answered the question. Taking $(a_n)=\frac{1}{n^2}$ will give the same problem, replace it with $(a_n)=\frac{1}{n^3}$ :). $\endgroup$ Jun 27, 2022 at 19:56

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