Let $X$ be a Banach space and $(x\_{n})\_{n=1}^{\infty}$ be a $p-$summable sequence in $X$. My basic questions are the following:


1. For any $\epsilon>0$, is there a sequence $(\xi\_{n})\_{n=1}^{\infty}$ such that $1\leq \xi\_{n}\rightarrow \infty(n\rightarrow \infty)$ and $\|(\xi\_{n}x\_{n})\_{n=1}^{\infty}\|\_{p}\leq \|(x\_{n})\_{n=1}^{\infty}\|\_{p}+\epsilon$;
2. Is there a sequence $(\xi\_{n})\_{n=1}^{\infty}$ such that $1\leq \xi\_{n}\rightarrow \infty(n\rightarrow \infty)$, $\sum\_{n=1}^{\infty}\frac{1}{\xi\_{n}^{p}}<\infty$ and $(\xi\_{n}x\_{n})\_{n=1}^{\infty}$ is a $p-$summable sequence in $X$?



Two basic questions on $p-$summable sequences
I assume "$p$-summable" means "absolutely $p$-summable", i.e., $\sum \|x\_n\|^p < \infty$.
Since all that matters about $x\_n$ is its norm, the question reduces to the scalar case.


1. Yes. Find a strictly increasing sequence of indices $(n\_k)$, $k \geq 1$, such that $\sum\_{i \geq n\_k} \|x\_i\|^p < \frac{\epsilon}{2^k}$. Then define $\xi\_n = 1$ for $n < n\_1$ and $\xi\_n = (k+1)^{1/p}$ for $n\_k \leq n < n\_{k+1}$. Now we have $$\sum\_{n=1}^\infty \xi\_n^p\|x\_n\|^p = \sum\_{n=1}^\infty \|x\_n\|^p + \sum\_{k=1}^\infty\sum\_{i=n\_k}^\infty \|x\_i\|^p \leq \sum\_{n=1}^\infty \|x\_n\|^p + \epsilon,$$ which is good enough.
2. Not for general $(x\_n)$. For example, let $x\_n = n^{-2/p} \in \mathbb{R}$ and let $(\xi\_n)$ be any sequence of positive numbers. For each $n$ we either have $\xi\_n^p \geq n$ or $\xi\_n^p < n$. In the first case, $\frac{\xi\_n^p}{n^2} \geq \frac{1}{n}$, and in the second case, $\frac{1}{\xi\_n^p} > \frac{1}{n}$. So $\sum\big(\frac{1}{\xi\_n^p} + \frac{\xi\_n^p}{n^2}\big) = \infty$ and therefore either $\sum\frac{1}{\xi\_n^p} = \infty$ or $\sum \xi\_n^p x\_n^p = \sum\frac{\xi\_n^p}{n^2} = \infty$.


