$u_n=a( \frac{1}{n}+a)^{-1} $ is an approximate unit of $A$ if $a$ is strictly positive

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Let $A$ be a $C^\ast$-algebra, $a\in A$ such that $\varphi(a)>0$ for all states $\varphi$ on $A$. How to prove, that $(u_n)_{n\in\mathbb{N}}$ with $u_n=a( \frac{1}{n}+a)^{-1} $ is an approximate unit of $A$?

It is to show that $\|b-u_nb\|\to 0$ and $\|b-bu_n\|\to 0$ for all $b\in A$, $n\to \infty$.

The only thing I know is that for nondegenerate $*$-representations $\pi:A\to L(H)$, $id_H$ is the strong operatortopology-limit of $(\pi(u_n))_{n\in\mathbb{N}}$. But I don't have a real idea how to use it or how to prove that $(u_n)$ is an approximate unit of $A$.

Do you have an idea or a hint for me?

Regards

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You prove this as I mentioned in my comment to your other question. You look at $(u_n)$ in the unitization $\bar A$ (note that you need the unitization in any case, to consider $(1/n+a)^{-1}$).

In the unital $\bar A$, you show using functional calculus that $a(1/n+a)^{-1}\to 1$: as $a$ is invertible in $\bar a$, its spectrum does not contain $0$, and this allows you to show that $t/(1/n+t)\to 1$ uniformly.

In all $u_n\to 1$ in $\bar A$, which implies that it is an approximate unit in $A$.