Suppose that $T:\ell_{2}\to \ell_{2}$ is an operator, such that $$(Tx)_{k} = \dfrac{k}{k+1}x_{k+1}$$ Find $\sigma(T)$.
Suppose that $|\lambda|>1$. Let $x_{k}\in \ell_{2}$. Then $$\dfrac{kx_{k+1}}{k+1}-\lambda x_{k}$$ yields (if I am not mistaken) $$x_{k} = \dfrac{k}{k-1}x_{k-1}+\dfrac{k}{k-2}x_{k-2}\lambda +\ldots+k\lambda^{k-2}x_1$$ But then $x\notin \ell_{2}$, so $T-\lambda I$ is not surjective, hence $\lambda\in \sigma(T)$ (recall that $|\lambda|>1$), which cannot be true, since $\sigma(T)$ is bounded. So, where is the mistake?
Your problem is taking $\ x_1=0\ $, whereas its value is in fact given by $$ x_1=-\sum_{i=1}^\infty\frac{y_i}{i\lambda^i}\ . $$ Including this initial condition in your recursion for $\ x_k\ $ gives \begin{align} x_k&=\lambda^{k-1}kx_1+k\sum_{j=1}^{k-1}\frac{\lambda^{k-j-1}y_j}{j}\\ &=-k\sum_{i=1}^\infty\frac{\lambda^{k-i-1}y_i}{i}+k\sum_{j=1}^{k-1}\frac{\lambda^{k-j-1}y_j}{j}\\ &=-k\sum_{j=k}^\infty\frac{\lambda^{k-j-1}y_j}{j}\\ &=-\sum_{i=0}^\infty\frac{ky_{i+k}}{\lambda^{i+1}(k+i)}\ , \end{align} and $\ x\ $, thus defined, is in $\ \ell_2\ $.
I found the value of $\ x_1\ $ (and $\ x_k\ $ too, as a matter of fact) by using the identity $$ (T-\lambda I)^{-1}=-\lambda^{-1}\sum_{i=0}^\infty\left(\frac{T}{\lambda}\right)^i $$ for $\ \lambda>1\ $. Offhand, I can't think of any other way of finding it.