Im attempting to multiply the infinite matrices $$A = \begin{bmatrix} 0 & \sqrt 1 & 0 & 0 & 0 & ... \\ \sqrt{1} & 0 & \sqrt{2} & 0 & 0 & ... \\ 0 & \sqrt 2 & 0 & \sqrt 3 & 0 & ... \\ 0 & 0 & \sqrt 3 & 0 & \sqrt 4 & ... \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix}$$ and $$B = \begin{bmatrix} 0 & - \sqrt 1 & 0 & 0 & 0 & ... \\ \sqrt{1} & 0 & - \sqrt{2} & 0 & 0 & ... \\ 0 & \sqrt 2 & 0 & - \sqrt 3 & 0 & ... \\ 0 & 0 & \sqrt 3 & 0 & - \sqrt 4 & ... \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix}$$ in index form. Multiplying in the traditional sense, I should end up with the infinite identity matrix. In index form, I write
$$A_{jk} = \sqrt k \delta_{j-1,k} + \sqrt j \delta_{j,k-1}$$ and $$B_{jk} = \sqrt k \delta_{j-1,k} - \sqrt j \delta_{j,k-1}$$ so the elements of the product $C := AB$ are given by $$C_{jk} = \sum_{\ell=1}^\infty A_{j \ell} B_{\ell k} = {\sum_{\ell=1}^\infty \Big( \sqrt{jk} \delta_{j,\ell-1}\delta_{\ell-1,k} - \sqrt{j\ell} \delta_{j,\ell-1}\delta_{\ell,k-1} + \sqrt{k \ell} \delta_{j-1,\ell}\delta_{\ell-1,k} - \sqrt{\ell^2} \delta_{j-1,\ell}\delta_{\ell,k-1}} \Big). $$ Then $$\sum_\ell \sqrt{jk} \delta_{j,\ell-1}\delta_{\ell-1,k} = \sqrt{jk}\delta_{j,k} = j\delta_{j,k}$$ $$\sum_{\ell} \sqrt{j\ell} \delta_{j,\ell-1}\delta_{\ell,k-1} =\sqrt{j(j+1)} \delta_{j,k-2}$$ $$\sum_\ell \sqrt{k \ell} \delta_{j-1,\ell}\delta_{\ell-1,k} = \sqrt{k(k+1)}\delta_{j-2,k}$$ $$\sum_\ell \sqrt{\ell^2} \delta_{j-1,\ell}\delta_{\ell,k-1} = (j-1)\delta_{j-1,k-1} $$ so $$C_{jk} = j\delta_{j,k}-\sqrt{j(j+1)} \delta_{j,k-2} + \sqrt{k(k+1)}\delta_{j-2,k}-(j-1)\delta_{j-1,k-1} \\ = \delta_{j,k}+\sqrt{k(k+1)}\delta_{j-2,k}-\sqrt{j(j+1)} \delta_{j,k-2}$$ which in principle should simplify to $\delta_{jk}$, however, I don't see how $\sqrt{k(k+1)}\delta_{j-2,k}=\sqrt{j(j+1)} \delta_{j,k-2}$. Did I make a mistake somewhere?