We consider $E_p=(c_{00}, ||\cdot||_{p})$ (where $c_{00}$ are the sequences that are zero except in a finite number of values and $1\leq p \leq \infty$) and the sequence:$$(x_n)_{n\in \mathbb{N}}=\big(\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}} , \stackrel{n^3}{\cdots},\frac{1}{\sqrt{n}}, 0,0,0, \cdots\big)$$ For which values of $p$ the sequence $(x_n)$:
a) $(x_n)$ is a bounded sequence?
b) $(x_n)$ is convergent in norm in $E_p$.
b) $(x_n)$ is weakly convergent in $E_p$.
My attempt (which I'm not sure if it's right) is the following:
a) As $\frac{1}{\sqrt{n}}\stackrel{n \rightarrow \infty}{\longrightarrow} 0 \Rightarrow (x_n)\stackrel{n \rightarrow \infty}{\longrightarrow}(0,0,0, \cdots), \forall p \in [1, \infty]$, then $(x_n)$ is bounded for all $p$.
b) Fixed $n \in \mathbb{N}$, then $(x_n)\stackrel{||\cdot||_p}{\longrightarrow} 0 \Leftrightarrow ||(x_n)||_p < \infty \Leftrightarrow ||(x_n)||_p= \big( n^3\cdot \big(\frac{1}{\sqrt{n}}\big)^{p} \big)^{1/p}=n^{(6p-1)/2p^2}< \infty \Leftrightarrow \frac{6p-1}{2p^2}<-1$
c) As convergence in norm implies weak convergence, all the $p$'s in $b)$ satisfy $c)$.
I don't know how to continue and if what I've done is right, so I'd appreciate any help. Thank you!
Promoting comment to answer:
Both parts (a) and (b) are about $\|(x_n)\|$. In the first, you should calculate this norm and find when it's bounded. In the second, when it goes to zero. Keep in mind that these are sequences, not series. The sequence $n^p$ is bounded iff $p\le 0$, and goes to zero iff $p<0$. Your "$<-1$" comes from thinking about series. For (c): recall that a sequence converges to 0 weakly iff the norm is bounded, and each coordinate goes to zero.