If $(\lambda_n)_{n\in\mathbb{N}}\in\ell^2$ then $(\lambda_{n+1}^2 + |\lambda_{n+1}-\lambda_n|)_{n\in\mathbb{N}}\in\ell^1$

Question

Let $(\lambda_n)_{n\in\mathbb{N}}\in\ell^2$, ie $\sum\limits_{n\in\mathbb{N}}|\lambda_n|^2<\infty$. Consider the sequence $(\lambda_n^2 + |\lambda_{n-1}-\lambda_n|)_{n\in\mathbb{N}}$. I wan to show that $(\lambda_{n+1}^2 + |\lambda_{n}-\lambda_{n+1}|)_{n\in\mathbb{N}}\in\ell^1$. My attempt is as follows:

$\sum\limits_{n\in\mathbb{N}}|\lambda_{n+1}^2 + |\lambda_{n+1}-\lambda_n|| = \sum\limits_{n\in\mathbb{N}}\lambda_{n+1}^2 + |\lambda_{n+1}-\lambda_n| =\sum\limits_{n\in\mathbb{N}}\lambda_{n+1}^2 + \sum\limits_{n\in\mathbb{N}}|\lambda_{n+1}-\lambda_n|\leq \sum\limits_{n\in\mathbb{N}}\lambda_{n+1}^2+\sum\limits_{n\in\mathbb{N}}|\lambda_{n+1}|+\sum\limits_{n\in\mathbb{N}}|\lambda_{n}|<\infty$

Since $(\lambda_n)_{n\in\mathbb{N}}\in\ell^2\supset\ell^1$ and so each of the three sums above is finite.

Answer 1

If $ \lambda_n=\frac{(-1)^n}{n}$, then $\lambda_n^2 + |\lambda_{n-1}-\lambda_n| \ge \frac{1}{n}$. hence

$(\lambda_{n+1}^2 + |\lambda_{n}-\lambda_{n+1}|)_{n\in\mathbb{N}}\notin\ell^1$.