Why is this a bounded operator? Let $\mathcal{H}$ be the Hilbert space $l^2(\mathbb{N})\otimes l^2(\mathbb{Z})$. I want to prove that the operator $T$ defined by 
$$T:=\sum_{k=1}^{\infty}{\sqrt{1-q^{2k}}e_{k-1,k}\otimes 1}$$
is a bounded operator on $\mathcal{H}$. Here $e_{k-1,k}$ is the standard matrix unit and $0<q<1$. One can prove that $||e_{k-1,k}||\leq1$. From this I did the following:
\begin{align*}
||T||&\leq\sum_{k=1}^{\infty}{||{\sqrt{1-q^{2k}}e_{k-1,k}\otimes 1}||}\\
&=\sum_{k=1}^{\infty}\sqrt{1-q^{2k}}||e_{k-1,k}||\cdot||1||\\
&\leq\sum_{k=1}^{\infty}\sqrt{1-q^{2k}}
\end{align*}
Therefore it suffices to prove that the sum converges, but from my point of view it doesn't. Maybe I did some mistakes above. Can someone help me? 
Thanks a lot.
 A: It is a general fact that the tensor product of bounded operators gives a bounded operator. Since your operator is of the form $S\otimes 1$, it is enough to show that $S$ is bounded. As it is defined in terms of a series, we need to check that such series converges. 
Write $q_k=\sqrt{1-q^{2k}}$. Note that $0\leq q_k\leq 1$ (assuming $|q|\leq1$). Given any $\xi=\sum_nc_n\delta_n\in\ell^2(\mathbb N)$, we have
$$
\left\|\sum_{k=r}^mq_k e_{k-1,k}\xi\right\|^2=\left\|\sum_{k=r}^m\sum_{n=1}^\infty q_kc_n e_{k-1,k}\delta_n\right\|^2
=\left\|\sum_{k=r}^m q_kc_k \delta_{n-1}\right\|^2=\sum_{k=r}^m|q_kc_k|^2\leq\sum_{k=r}^m|c_k|^2.
$$
As the sequence $\{c_k\}$ is in $\ell^2(\mathbb N)$ (because $\xi\in\ell^2(\mathbb N)$ ), this shows that the sequence of partial sums defining $S$ is Cauchy, and so the series exists. The same estimate above, starting with $r=1$, shows that $\|S\|\leq1$. 
Finally, using the fact that the spatial tensor norm on bounded operators is a cross norm, $\|T\|=\|S\otimes 1\|=\|S\|$. 
