Proving that $\Big(\sum_{k=1}^\infty a_k\Big)^p\leq c \sum_{k=1}^\infty \beta^ka_k^p.$ Let  $p>0$ and $\beta>1$ then there exists a constant $c=c(p,\beta)$ such for every sequence $(a_k)_k$ with $a_k\geq0$ we have
$$\Big(\sum_{k=1}^\infty a_k\Big)^p\leq c \sum_{k=1}^\infty \beta^ka_k^p$$
The case $p\leq 1$ is obvious here since we have $(a_1+a_2)^p\leq a_1^p+a_2^p$ and by induction one easily arrives at
$$\Big(\sum_{k=1}^\infty a_k\Big)^p\leq  \sum_{k=1}^\infty a_k^p\leq \sum_{k=1}^\infty \beta^ka_k^p$$
How to prove the case $p>1$?
 A: I consider the case that $p>1$ and $\beta>1$ and the rest is left
to you. We need some preparation.
Let $\gamma\in(0,1)$. Let $p>1$. Let $q\in(1,\infty)$ be such that
$\frac{1}{p}+\frac{1}{q}=1$. Define $T:l^{p}\rightarrow l^{1}$ by
$Ta=(a_{1}\gamma,a_{2}\gamma^{2},\ldots),$ where $a=(a_{1},a_{2},\ldots)\in l^{p}$.
Firstly, we show that $T$ is well-defined. By Holder inequality,
\begin{eqnarray*}
\sum_{k=1}^{\infty}|a_{k}\gamma^{k}| & \leq & \left\{ \sum_{k=1}^{\infty}|a_{k}|^{p}\right\} ^{\frac{1}{p}}\left\{ \sum_{k=1}^{\infty}\gamma^{kq}\right\} ^{\frac{1}{q}}\\
 & = & M||a||_{p},
\end{eqnarray*}
where $M=\left\{ \sum_{k=1}^{\infty}\gamma^{kq}\right\} ^{\frac{1}{q}}<\infty$.
Note that $M$ depends on $\gamma$ and $p$ only. It is easy to show
that $T$ is linear and $||T||\leq M$.
Now, we go back to your question. Let $p>1$ and $\beta>1$ be given.
We define $\beta_{0}=\beta^{\frac{1}{p}}>1$, $\gamma=\frac{1}{\beta_{0}}$,
and $M=\left\{ \sum_{k=1}^{\infty}\gamma^{kq}\right\} ^{\frac{1}{q}}<\infty$.
Let $a=(a_{n})$ be given with $a_{n}\geq0$. For each $n\in\mathbb{N}$,
define $$b^{(n)}=(a_{1}\beta_{0},a_{2}\beta_{0}^{2},\ldots,a_{n}\beta_{0}^{n},0,0,\ldots).$$
Clearly $b^{(n)}\in l^{p}$. It follows that $||Tb^{(n)}||_{1}\leq M||b^{(n)}||_{p}$.
That is,
\begin{eqnarray*}
\sum_{k=1}^{n}a_{k} & \leq & M\left\{ \sum_{k=1}^{n}a_{k}^{p}\beta_{0}^{kp}\right\} ^{\frac{1}{p}}\\
 & = & M\left\{ \sum_{k=1}^{n}a_{k}^{p}\beta^{k}\right\} ^{\frac{1}{p}}\\
 & \leq & M\left\{ \sum_{k=1}^{\infty}a_{k}^{p}\beta^{k}\right\} ^{\frac{1}{p}}.
\end{eqnarray*}
Letting $n\rightarrow\infty$ yields $\sum_{k=1}^{\infty}a_{k}\leq M\left\{ \sum_{k=1}^{\infty}a_{k}^{p}\beta^{k}\right\} ^{\frac{1}{p}}$.
A: Let $p>1$, and let $q=\frac{p}{p-1}$ be the Lebesgue conjugate exponent of $p$, and let $\beta>1$.
First write
$$\sum_{k=1}^{\infty} a_k= \sum_{k=1}^{\infty}\frac{1}{\beta^{\frac{k}{p}}} \beta^{\frac{k}{p}} a_k.$$
Then apply  Holder's inequality to get
$$\sum_{k=1}^{\infty} a_k\leq \left(\sum_{k=1}^{\infty}\left(\frac{1}{\beta^\frac{q}{p}}\right)^k\right)^{\frac{1}{q}} \left(\sum_{k=1}^{\infty} \beta^{k} a_k^p\right)^{\frac{1}{p}}.$$
The sum $\sum_{k=1}^{\infty}\left(\frac{1}{\beta^\frac{q}{p}}\right)^k$ converges since $\beta^\frac{q}{p}>1$.
