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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?

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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 $$2^{1+\frac{2}{p}} = 4 \implies 1+\frac{2}{p} = 2 \implies p = 2$$

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