I want to prove the following theorem:
Let $1\le p < \infty$ and $K\subset \ell^p$. If $\forall n\in\mathbb{N},\ \exists M_n>0$, such that $\sup_{a\in K} |a_n|<M_n$ and $\forall \varepsilon > 0, \ \exists N(\varepsilon)$ with $\sum_{n=N+1}|a_n|^p\le \varepsilon^p$ for all $a\in K$, then $K$ is relatively compact.
My attempt:
Since the space $\ell^p$ is normed, a metric can be induced. Furthermore, I know that in metric spaces relatively compactness is equal to total boundedness. Therefore, I tried to show total boundedness of $K$. Hence I have to obtain a finite cover of epsilon balls for arbitrary epsilon.
For fixed $a'\in K$ and any $x\in K$ it holds that,
$\| x-a'\|_{\ell^p} \le \|x\|_{\ell^p} + \|a'\|_{\ell^p}= \left ( \sum_{n=0}^\infty |x_n|^p\right )^{1/p} + \left (\sum_{n=0}^\infty |a_n|^p\right )^{1/p}=\left ( \sum_{n=0}^{N(\varepsilon)}|x_n|^p+ \sum_{n={N(\varepsilon)}+1}^\infty|x_n|^p\right )^{1/p}+\left ( \sum_{n=0}^{N(\varepsilon)}|a_n|^p+ \sum_{n={N(\varepsilon)}+1}^\infty|a_n|^p\right )^{1/p}$
(I used the Minkowski-Inequality). I define $\sum_{i=0}^{N(\varepsilon)} M_j =\mu({N(\varepsilon)})$. It then follows from the assumptions in the theorem that $\| x-a'\|_{\ell^p}\le 2\left (\mu(N(\varepsilon)) + \epsilon^p\right )^{1/p}$
This means that I can take a ball of radius $2\left (\mu(N(\varepsilon)) + \epsilon^p\right )^{1/p}$ with arbitrary $a'\in K$ and this for any given $\varepsilon>0$.
I have my doubts, since I can cover $K$ for a given $\varepsilon$ with only one ball and this for every $a'$.
