A class of sequences is of bounded variation Let $\left\{ a_n \right\}$ be a null sequence s.t. 
$$\sum_{n=1}^\infty \left( \frac{1}{n}\sum_{k=n}^\infty \vert\Delta a_k\vert^p\right)^\frac{1}{p}<\infty$$
for some $p>1$.
How to prove that $\left\{ a_n \right\}$ must be of bounded variation?
My attempt: It is enough to prove that
$$\vert \Delta a_n\vert \leq \left( \frac{1}{n}\sum_{k=n}^\infty \vert\Delta a_k\vert^p\right)^\frac{1}{p}, \forall n>n_0, \text{for some} \; n_0 \in \mathbb{N}.$$
I've tried proving the last inequality using Jensens' inequality, but I can't get to the desired result.
Edit:
How to find a sequence $\left\{ a_n \right\}$ of bounded variation for which $\sum_{n=1}^\infty \left( \frac{1}{n}\sum_{k=n}^\infty \vert\Delta a_k\vert^p\right)^\frac{1}{p}<\infty$ doesn't hold for any $p>1$?
 A: Let, as usual, $q$ denote the conjugate exponent to $p$. For each $n$, by Hölder's inequality we have
$$\sum_{k = n}^{2n-1} \lvert \Delta a_k\rvert = \sum_{k = n}^{2n-1} 1\cdot \lvert\Delta a_k\rvert \leqslant n^{1/q}\biggl(\sum_{k = n}^{2n-1} \lvert\Delta a_k\rvert^p\biggr)^{1/p},$$
that is,
$$\frac{1}{n}\sum_{k = n}^{2n-1} \lvert \Delta a_k\rvert \leqslant \biggl(\frac{1}{n}\sum_{k = n}^{2n-1} \lvert\Delta a_k\rvert^p\biggr)^{1/p} \leqslant \biggl(\frac{1}{n}\sum_{k = n}^{\infty} \lvert\Delta a_k\rvert^p\biggr)^{1/p}.$$
Consequently
$$\sum_{n = 1}^{\infty} \frac{1}{n}\sum_{k = n}^{2n-1} \lvert\Delta a_k\rvert < \infty.$$
Now we can change the order of summation and obtain
$$\sum_{k = 1}^{\infty} \lvert\Delta a_k\rvert \sum_{k/2 < n \leqslant k} \frac{1}{n} < \infty.$$
And for every $k$ we have
$$\sum_{k/2 < n \leqslant k} \frac{1}{n} \geqslant \frac{1}{2}\,,$$
hence
$$\sum_{k = 1}^{\infty} \lvert\Delta a_k\rvert \leqslant 2\sum_{n = 1}^{\infty} \frac{1}{n}\sum_{k = n}^{2n-1} \lvert\Delta a_k\rvert < \infty\,,$$
i.e. $(a_k)$ is of bounded variation.
With regard to the edit, note that
$$\biggl(\frac{1}{n}\sum_{k = n}^{\infty} \lvert\Delta a_k\rvert^p\biggr)^{1/p} \geqslant \frac{1}{n^{1/p}}\max \{ \lvert\Delta a_k\rvert : k \geqslant n\} \geqslant \frac{1}{n}\max \{ \lvert\Delta a_k\rvert : k \geqslant n\}\,.$$
Thus from any null sequence $(b_n)$ of bounded variation for which $\Delta b_n \neq 0$ infinitely often we can construct a null sequence $(a_n)$ of bounded variation with the desired property by choosing a sufficiently fast (what is sufficiently fast depends on $(b_n)$) increasing sequence $(n_m)$ of positive integers and setting $a_n = b_m$ for $n_m < n \leqslant n_{m+1}$. Then
$$\sum_{n = n_{m-1}+1}^{n_m} \biggl(\frac{1}{n}\sum_{k = n}^{\infty} \lvert\Delta a_k\rvert^p\biggr)^{1/p} \geqslant \sum_{n = n_{m-1}+1}^{n_m} \frac{1}{n}\lvert\Delta a_{n_m}\rvert \sim \lvert\Delta b_m\rvert\log \frac{n_m}{n_{m-1}}$$
and we can choose $(n_m)$ such that
$$\sum_{m = 2}^{\infty} \lvert\Delta b_m\rvert\log \frac{n_m}{n_{m-1}} = +\infty\,.\tag{$\ast$}$$
As an example, let $b_m = \frac{1}{m}$ and $n_m = 2^{m(m-1)}$. Thus $\Delta b_m = \frac{1}{m(m+1)}$ and $\log (n_m/n_{m-1}) = (m-1)\log 4$, and $(\ast)$ diverges by comparison to the harmonic series.
A: The argument is:
We wish to prove $n|\Delta a_n|^p \leq \sum_{k=n}^{\infty} |\Delta a_k|^p$.
Now suppose that the right hand side does not contain at least $n$ terms of size $|\Delta a_n|^p$. Then surely $|\Delta a_k|^p \leq |\Delta a_n|^p$ eventually (i.e. of BV). If it does contain at least $n$ terms of this size, then the inequality holds, and again $a_n$ is of bounded variation as well.
