What is the norm of the operator $L((x_n)) \equiv \sum_{n=1}^\infty \frac{x_n}{\sqrt{n(n+1)}}$ on $\ell_2$?

Let $(x_n) \subset \ell_2$ and let operator $L:\ell_2\to \mathbb R$ be defined by:

$\displaystyle L((x_n)) := \sum_{n=1}^{\infty} \frac{x_n}{\sqrt{n(n+1)}}$.

Find the norm of L.

Solutions Collecting From Web of "What is the norm of the operator $L((x_n)) \equiv \sum_{n=1}^\infty \frac{x_n}{\sqrt{n(n+1)}}$ on $\ell_2$?"

Let $a\in\ell^2$ the sequence defined by $a(n):=\frac 1{\sqrt{n(n+1)}}$. Then $T(x)=\langle a,x\rangle_{\ell^2}$.

Cauchy-Schwarz inequality gives that $$\lVert T\rVert=\lVert a\rVert_{\ell^2}= \sum_{n=1}^{+\infty}\frac{n+1-n}{n(n+1)}=\sum_{n=1}^{+\infty}\left(\frac 1n-\frac 1{n+1}\right)=1.$$

$$L(x) = \left\langle x,\Bigl((n(n+1))^{-1/2}\Bigr)\right\rangle = \langle x,\xi\rangle.$$

So by the Riesz representation theorem for Hilbert spaces we have

$$\lVert L\rVert^2 = \lVert \xi\rVert_2^2 = \sum_{n=1}^\infty \left|\frac1{\sqrt{n(n+1)}}\right|^2 = \sum_{n=1}^\infty \frac1{n(n+1)}.$$

You should be able to sum the series yourself.