Strict convexity of $c_0$ Let $c_0$ be a spaces of sequences converging to $0$ with the following norm
$$
\|x\|=\sup\{|x_i|: i\in \mathbb{N}\}+\left(\sum_{i=1}^{\infty}\left(\frac{x_i}{i}\right)^2\right)^{\frac{1}{2}}
$$
Prove that $(c_0,\|.\|)$ is strictly convex but not uniformly convex. 
Thank you for your kind help.
I can prove that $(c_0,\|.\|)$ is not uniformly convex by choosing $\{x^k\}, \{y^k\}\subset c_0$ given by
$$
x^k_i=
\begin{cases}
\frac{1}{2}\sqrt{1-\frac{1}{(k+1)^2}-\frac{1}{(k+2)^2}}& \text{if}\quad i=1\\
\frac{1}{2}& \text{if}\quad i\in\{k+1, k+2\} \\
0&\text{if}\quad\text{otherwise}
\end{cases} 
$$
$$
y^k_i=
\begin{cases}
\frac{1}{2}\sqrt{1-\frac{1}{k^2}-\frac{1}{(k+2)^2}}& \text{if}\quad i=1\\
\frac{1}{2}& \text{if}\quad i\in\{k, k+2\} \\
0&\text{if}\quad\text{otherwise}
\end{cases} 
$$
We are easy to check $\|x^k\|=\|y^k\|=1$, $\|x^k-y^k\|\geq \frac{1}{2}$ but
$$
\left\|\frac{x^k+y^k}{2}\right\|\longrightarrow 1 \quad\text{as}\quad k\longrightarrow \infty.
$$

Following the guidance of Robert Israel on the link
Strict convexity of a norm on $C[0,1]$
we can prove that $(c_0, \|.\|)$ is strictly convex. Indeed, suppose that $x, y\in c_0\setminus\{0\}$ such that $\|x+y\|=\|x\|+\|y\|$. 
Consider the inner product and the norm on $c_0$ given by
$$
\langle x,y\rangle=\sum_{i=1}^{\infty}\frac{x_iy_i}{i^2},
$$
$$
\|x\|_1=\sqrt{\langle x,x\rangle}.
$$
Since $\|x+y\|=\|x\|+\|y\|$, it follows that $\|x+y\|_1=\|x\|_1+\|y\|_1$
If $x\ne\lambda y$ for all $\lambda\in\mathbb{R}$ then
$$
\langle x-\lambda y, x-\lambda y \rangle>0,
$$
$$
\langle x+\lambda y, x+\lambda y \rangle>0
$$
for all $\lambda\in\mathbb{R}$.
Expanding the following inequalities we obtain $|\langle x, y\rangle|<\|x\|\|y\|$
It follows that
$\|x+y\|_1<\|x\|_1+\|y\|_1$, an absurd.
