# Let $z_n>0$ and $\liminf z_n=0$. Show that exists $x_n>0$ such that $\sum x_n<\infty$ and $\limsup \frac{x_n}{z_n}=\infty$

Let $z_n>0$ and $\liminf z_n=0$. Show that exists $x_n>0$ such that $\sum x_n<\infty$ and $\limsup \frac{x_n}{z_n}=\infty$

I need some help with this proof.

We need some $(x_n)$ positive such that it series $\sum x_n\to x$ converges, what implies that $(x_n)\to 0$.

If $\limsup \frac{x_n}{z_n}=\infty$ then exists some subsequences $(x_{n_k})$ and $(z_{n_k})$ such that $\left(\frac{x_{n_k}}{z_{n_k}}\right)\to\infty$. But $(x_{n_k})\to 0$ because $(x_n)\to 0$. Then we need to set some $(x_{n_k})\to 0$ slowly enough such that $\left(\frac{x_{n_k}}{z_{n_k}}\right)\to\infty$.

Because $\liminf z_n=0$ we know that exists some subsequence $(z_{n_k})\to 0$. Then we pick a subsequence of $(z_n)$ and we set $x_{n_k}:=\sqrt{z_{n_k}}$. Then obviously we have that

$$z_{n-k}>0\land (z_{n_k})\to 0\implies (z_{n_k}^{-\frac12})\to\infty$$

But I dont know how to ensure that exists $\sum x_{n_k}<\infty$ for some $(z_{n_k})$. I thought about a criteria to choose, if needed, other subsequence of the subsequence $(x_{n_k})$ such that it series will be convergent (after we can fill the gaps with a constant sequence enough small that $\sum x_n$ converges), but I cant find a clever criteria to write this proof correctly.

UPDATE: I think that I know how to do it... we can choose a convergent series as criteria, by example $\sum\frac1{n^4}$.

Then, for a subsequence $(z_{n_k})\to 0$ we filter this subsequence with the criteria $z_{n_k}\le \frac1{n_k^4}$ (we can do this because the neighborhood of zero have arbitrarily small $z_{n_k}$). Then we produce a second subsequence $(z_{n_k}')$ that ensures that $\sum x_{n_k}<\infty$. From here we can fill the sequence $x_{n_k}$ to produce $x_n$ just with $\frac1{n^2}$ in the holes.

It is correct this idea? If it is, there is a way to write this formally.

Since $\liminf_{n\rightarrow \infty} z_n = 0$, for each natural number $k$ we may find $z_{n_k}$ such that $z_{n_k}<\frac{1}{k^3}$. We may assume $n_1<n_2<\cdots$. Define $x_{n_k}=kz_{n_k}$ and $x_i=\frac{1}{2^{i}}$ for $i\notin \{n_1,n_2,\ldots\}$. Then $x_n>0$ and $$\sum_{n} x_n \leq \sum_{i>0} \frac{1}{2^{i}} + \sum_{k>0} \frac{1}{k^2} < \infty,$$ and also $$\lim_{k\rightarrow \infty} \frac{x_{n_k}}{z_{n_k}}=\lim_{k\rightarrow \infty} k = \infty,$$ so $x_n$ is as required.
• Clean answer. I think just saying $x_i = 0$ for $i \not\in \{n_1, \dots \}$ is sufficient too. – Adam Jaffe Oct 28 '16 at 1:27
• It was required that $x_i>0$. – SMJK Oct 28 '16 at 1:34