# $l^p$ space not having inner product

I know that $l^2$ space is a Hilbert space. But for other $l^p$ spaces, where $p\geq1$, I have to show that they do not satisfy the parallelogram equality.

But, I can't find appropriate sequences that become counterexamples of the parallelogram equality. Could anyone suggest me some sequences?

Consider $\mathbf{x} = (1,0,0,0,\ldots)$ and $\mathbf{y} = (0,1,0,0,\ldots)$.
Then $\mathbf{x+y} = (1,1,0,0,\ldots)$ and $\mathbf{x-y} = (1,-1,0,0,\ldots)$.
$$\|\mathbf{x}+\mathbf{y}\|_p^2 + \|\mathbf{x}-\mathbf{y}\|_p^2 = 2\times(1+1)^\frac{2}{p} = 2^{1+\frac{2}{p}}$$
Meanwhile, $$2(\|\mathbf{x}\|_p^2+\|\mathbf{y}\|_p^2) = 2\times(1+1) = 4$$
If the parallelogram law holds, then $$2^{1+\frac{2}{p}} = 4 \implies 1+\frac{2}{p} = 2 \implies p = 2$$