Operator norm of $T:l^{2}\rightarrow l^{1}$ where $Tx=(x_{1},x_{2}/2,x_{3}/3,x_{4}/4,...)$ As the title states, I need to compute the operator norm of a linear operator 
$T:l^{2}\rightarrow l^{1}$, where 
$$Tx=\left(x_{1},\frac{x_{2}}{2},\frac{x_{3}}{3},\frac{x_{4}}{4},... \right)$$
Using Holder's inequality for any sequence $(x_{i})_{i\geq 1}\in l^{2}$, we can show
\begin{align}
|Tx|_{1}&=\sum_{i=1}^{\infty}=|x_{1}|+\left|\frac{x_{2}}{2}\right|+\left|\frac{x_{3}}{3}\right|+\cdots\\
&=|x_{1}||1|+|x_{2}|\left|\frac{1}{2}\right|+|x_{3}|\left|\frac{1}{3}\right|+\cdots \\
&\leq \left|x_{i}\right|_{2}\left|\frac{1}{i}\right|_{2} \\
&=\frac{\pi}{\sqrt{6}}|(x_{i})|_{2}
\end{align}
Hence 
$$\displaystyle |Tx|_{1}\leq\frac{\pi}{\sqrt{6}}|(x_{i})|_{2}\implies||T||\leq\frac{\pi}{\sqrt{6}}$$
However, I am unable to find a sequence in $l_{2}$ which has norm $|(x_{i})|\leq 1$ so that I may use the property $||T||=\text{sup}_{|(x_{i})|_{2}=1}|Tx|_{1}$.
Any help is appreciated. Thank you.

Let $x_i=\frac c i, i=1,2\cdots$ where $c$ is such that $c\sum\limits_{i=1}^{\infty} x_i^{2}=1$. In other words, $c=\frac {\sqrt 6} {\pi}$. Then $\|T(x_i)\|=\sum\limits_{i=1}^{\infty} \frac c {i^{2}}$ which is exactly $\frac {\pi} {\sqrt 6}$.
