proof of an equality norm Let the mapping $T:\ell^{2}\rightarrow \ell^{2}$ is defined as follow.
$$T(x_1,x_2,\ldots,x_n,\ldots)=(x_1,\dfrac{1}{2}x_2,\ldots,\dfrac{1}{n}x_n,\ldots)$$
In this case, i've easily earned:
$$\sigma(T)=\{0\}\cup\{\dfrac{1}{n}:n\in\mathbb{N}\}$$
Now, if $\lambda=x+iy\in\mathbb{C}$ and $\lambda\notin\sigma(T)$, i should prove
$$\Vert(\lambda I-T)^{-1}\Vert=\dfrac{1}{\displaystyle\inf_{n\in\mathbb{N}}\,\left\vert\lambda-\dfrac{1}{n}\right\vert}$$
can you help me for starting of prove?
Thanks.
 A: You have
$$
(\lambda I-T)(x_1,x_2,\ldots)=((\lambda-1)x_1, (\lambda-\frac12)x_2,\ldots). $$
Define an operator  $S $ by $$S (x_1,x_2,\ldots)=(\frac{x_1} {\lambda -1},\frac {x_2}{\lambda -\frac12},\ldots). $$ By the choice of $\lambda  $ the linear operator $S $ is well-defined and bounded: by construction, $S (\lambda I-T)=(\lambda I-T)S=I $. So $S=(\lambda I-T)^{-1} $ (bounded, because $\lambda\not\in\sigma (T) $. And $$\tag{1}\|S\|=\sup\left\{\frac1 {\left|\lambda-\frac1n\right|}:\ n\in\mathbb N\right\}=\frac1 {\inf\left\{\left|\lambda-\frac1n\right |:\ n\in\mathbb N\right\}}. $$ 
$\\ \ $

The only piece left hanging, to justify the first equality in $(1)$, is to show that if $$ R (x_1,x_2,\ldots)=(r_1x_1,r_2x_2,\ldots) $$ then $\|R\|=\sup\{|r_n|:\ n\} $. This follows, if we write $c=\sup\{|r_n|:\ n\} $:
$$
\|Rx\|^2=\sum_j|r_jx_j|^2\leq c^2\sum_j|x_j|^2=c^2\,\|x\|^2,
$$
so $\|R\|\leq c$. Let $j$ such that $|r_j|>r-\frac1j$. If $e_j\in\ell^2$ is the sequence with a $1$ in the $j^{\rm th}$ position and zeroes elsewhere, then $\|e_j\|=1$ and 
$$
\|Re_j\|=|r_j|>r-\frac1j.
$$ So $\|R\|=c=\sup\{|r_n|:\ n\}$. 
