Consider an infinitely-lived agent’s consumption-saving problem. The agent receives $e > 0$ units of endowment every period, can save via an asset with constant return $R$. The agent is endowed with $s_0$ units of the asset initially. At period $t$, he chooses the amount to consume $c_t$ and to save $s_{t+1}$. At $t + 1$, his total available resources are endowment e and total returnon saving $R s_{t+1}$. The agent maximizes discounted lifetime utility $\sum_{t=0}^\infty \beta^t u(c_t)$ with discount factor $\beta\in (0, 1)$. He can not borrow, i.e., $s_{t+1} \geq 0$ for all $t ≥ 0$. Assume that $R \in [0, \frac{1}{\beta}]$, and u is strictly increasing, strictly concave, and continuously differentiable.
Suppose that $s_0=0$. Show that $s_{t+1}^*=0$ for all $t\geq 0$ is the unique solution to the sequential problem.
Here is my attempt:
The sequential problem is \begin{align} \max \sum_{t=0}^\infty \beta^t u(e+Rs_t-s_{t+1})\\ s.t. s_{t+1}\in\Gamma(s_t)\\ s_0\text{ is given} \end{align}
Suppose $s_0=0$. If we want to show $s_{t+1}^*=0$ for all $t\geq 0$. By induction hypothesis, let $s_t=0$. Thus, the sequential problem at period $t$ is \begin{align} \max\sum_{\tau=t}^\infty \beta^t u(c_\tau)\\ s.t. c_t+s_{t+1}=e\\ c_{\tau}+s_{\tau+1}=Rs_{\tau}+e\\ s_{t+1}\geq 0 \end{align}
Then the Lagrangian: $$\sum_{\tau=t}^\infty \beta^t u(c_\tau)+\lambda_t (e-s_{t+1}-c_t)+\lambda_{\tau}(Rs_{\tau}+e-c_{\tau}-s_{\tau+1})+\mu s_{t+1}$$
First order conditions: \begin{align} [s_{t+1}]\lambda_t=\lambda_{t+1}R+\mu\\ [c_{t}] \beta^t u'(c_t)=\lambda_t\\ [c_{t+1}]\beta^{t+1}u'(c_{t+1})=\lambda_{t+1} \end{align}
Suppose by contrdiction, $s_{t+1}^*\neq 0\implies s_{t+1}^*>0$. By complementary slackness, $\mu=0\implies \lambda_t=\lambda_{t+1}R\implies u'(c_t)=\beta u'(c_{t+1})R$. Since $\beta R\leq 1$, we have $u'(c_t)\leq u'(c_{t+1})\implies c_t\geq c_{t+1}$.
Since $s_{t+1}^*>0$ and $s_t=0$, by $c_t+s_{t+1}=Rs_t+e$, we have $c_t<e$. Thus, $c_{t+1}\leq c_t<e$.
I don't know how to derive a contradiction up. Can someone help?
