# Let $(a_n)$ be a sequence of positive real numbers, then which of the following is/are correct?

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.

• 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. Jun 27, 2022 at 6:09
• Thanks @QC_QAOA it discarded (3) and (1) both but (2) is the only correct answer. Any idea for (4)? Jun 27, 2022 at 8:56

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

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

• 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. Jun 27, 2022 at 19:35
• 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. Jun 27, 2022 at 19:44
• 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 $(a_n)=\frac{1}{n^3}$ :). Jun 27, 2022 at 19:56