Infinite amount of numbers such that $x_n > 1/n$. Then $\sum_n x_n$ doesn't converge (Edit: Here $\mathbb{R}_0$ means $\mathbb{R} \setminus \left\{0 \right\}$. Same for $\mathbb{N}_0$.) 
I need to prove the following:
Problem: Let $(x_n)_{n \in \mathbb{N}_0}$ be a decreasing sequence in $\mathbb{R}_0$. Suppose there exist an infinite amount of numbers $n \in \mathbb{N}_0$ such that $x_n > \frac{1}{n}$. Prove that $\sum_{n=1}^{\infty} x_n$ doesn't converge.
Attempt: Let $$s_n = \sum_{k = 1}^n = x_k$$ be the $n$th partial sum. Then we have $s_n - s_{n-1} = x_n$. There exist an infinite amount of $n \in \mathbb{N}_0$ such that $x_n > 1/n$. Because $\lim_{n \to \infty} 1/n = 0$, it follows by taking the limit that $$ \lim_{n \to \infty} s_n - \lim_{n \to \infty} s_{n-1} = \lim_{n \to \infty} x_n \geq 0. $$ Now I want to conclude from this somehow that $\lim_{n \to \infty} x_n \neq 0$, and hence the series is not convergent (because the terms don't go to zero). Would this be correct reasoning? 
 A: If $\sum_{n=1}^\infty x_n$ converges, then we must have $\lim_{N\to\infty}\sum_{n=N+1}^\infty x_n=0.$ Choose $N$ so that $\sum_{n=N+1}^\infty x_n\lt\frac12.$ Choose $k\gt2N$ with $x_k\ge\frac1k.$ Then
$$\sum_{n=N+1}^\infty x_n\ge\sum_{n=N+1}^k x_n\ge\sum_{n=N+1}^k\frac1k=\frac{k-N}k\gt\frac{k-\frac12k}k=\frac12$$
contradicting the way $N$ was chosen.
A slightly more general formulation: if $\{x_n\}$ is a decreasing sequence of positive numbers, and if $\limsup_{n\to\infty}nx_n\gt0,$ then $\sum_{n=1}^\infty x_n$ diverges.
A: Let $n_k$ be the subsequence for which the condition holds.
Note that $\sum_n x_n > \sum_k (1 - {n_k \over n_{k+1}})$.
Let $r = \liminf_k {n_k \over n_{k+1}}$. Note that $r \le 1$.
If $r<1$, then for some $r < \rho <1$ and $k$ sufficiently large
we have $(1 - {n_k \over n_{k+1}}) > 1-\rho$ infinitely often, hence
the series is divergent. 
Hence $r=1$, and for some $k_0$ sufficiently large we have ${n_k \over n_{k+1}} > {1 \over 2}$ for $k \ge k_0$.
We now mimic the proof that $\sum_n { 1\over n}$ is divergent.
In particular, this shows that $n_{k_0+i} < 2^i n_{k_0}$, and, in particular,
the range of indices
$2^{m}n_{k_0},...,2^{m+1}n_{k_0}-1$ must contain at least one index for which
$x_n > {1 \over n}$ and so
$\sum_{i=2^{m}n_{k_0}}^{2^{m+1}n_{k_0}-1} x_n > \sum_{i=2^{m}n_{k_0}}^{2^{m+1}n_{k_0}-1} {1 \over 2^{m+1}n_{k_0}} = {1 \over 2}$, and summing over a suitable range of $m$ shows that the series is
divergent.
