Intemporal budget by lagrange Assume that a representative agent lives forever and receives an endowment, denoted yt, in each period. The entire endowment sequence is known
with certainty on date 0. The representative agent discounts future utility at
rate $\beta$ and has a period utility function given by
$u(c_{t}) = (c_{t})^{1-\sigma}/(1-\sigma)  $

(d)write the intertemporal budget constraint of the representative
agent by aggregating the flow budget constraints.
Write down the Lagrangian using the constraint in (d). How many
Lagrange multipliers are there? Find the first order conditions and write down
the Euler equation.

I have found the equation 
$ L = \beta^tU(c_{t}) + (\lambda)_{t}(1/(1+r)^t)(c_{t}-y_{t}) $

then I have found the equation by taking derivative of L with respect to $c_{t}$ and $c_{t+1}$ but since there are no savings(b) I could not take derivative of L with respect to b so cannot find the ratio of lamdas :
$ \frac{U^{'}(c_{t})}{U^{'}(c_{t+1})\beta} = \frac{(\lambda)_{t}}{(\lambda)_{t+1}(r+1)}$
The lagrange-function should be look like this:

$\mathcal L=\sum_{t=0}^{n} \left( \beta ^t\cdot \frac{1}{1-\sigma}\cdot c_t^{1-\sigma}+ \lambda_t \left(  \frac{y_t}{(1+r)^t}- \frac{c_t}{(1+r)^t} \right) \right) $

$ y_t- c_t  =0\Rightarrow \text{No savings in period t}$

First order condition:

$\frac{d L}{d c_t}=\beta ^t\cdot \frac{1}{1-\sigma} \cdot (1-\sigma) c_t^{1-\sigma}-\frac{\lambda _t}{(1+r)^t}=0 \Rightarrow \lambda_t =\beta ^t\cdot \frac{(1+r)^t}{1-\sigma} \cdot (1-\sigma) c_t^{1-\sigma} \quad (1) $

$\frac{d L}{d c_{t+1}}=\beta ^{t+1}\cdot \frac{1}{1-\sigma} \cdot (1-\sigma) c_{t+1}^{1-\sigma}-\frac{\lambda _{t+1}}{(1+r)^{t+1}}=0 \Rightarrow \lambda_{t+1} =\beta ^{t+1}\cdot \frac{(1+r)^{t+1}}{1-\sigma} \cdot (1-\sigma) c_{t+1}^{1-\sigma} \quad (2) $

Dividing (2) by (1) gives $\frac{\lambda _{t+1} }{\lambda _{t}}=\beta \cdot (1+r) \frac{c_{t+1}^{1-\sigma}}{c_{t}^{1-\sigma}}$

It is the same as $\frac{\lambda _{t+1} }{\lambda _{t}}=\beta \cdot (1+r) \frac{U'(c_{t+1})}{U'(c_{t})}$

My relation is a little bit different from yours.
The lagrange-function should be look like this:

$\mathcal L=\sum_{t=0}^{n} \left( \beta ^t\cdot \frac{1}{1-\sigma}\cdot c_t^{1-\sigma}+ \lambda_t \left(  \frac{y_t}{(1+r)^t}- \frac{c_t}{(1+r)^t} \right) \right) $

$ y_t- c_t  =0\Rightarrow \text{No savings in period t}$

First order condition:

$\frac{d L}{d c_t}=\beta ^t\cdot \frac{1}{1-\sigma} \cdot (1-\sigma) c_t^{1-\sigma}-\frac{\lambda _t}{(1+r)^t}=0 \Rightarrow \lambda_t =\beta ^t\cdot \frac{(1+r)^t}{1-\sigma} \cdot (1-\sigma) c_t^{1-\sigma} \quad (1) $

$\frac{d L}{d c_{t+1}}=\beta ^{t+1}\cdot \frac{1}{1-\sigma} \cdot (1-\sigma) c_{t+1}^{1-\sigma}-\frac{\lambda _{t+1}}{(1+r)^{t+1}}=0 \Rightarrow \lambda_{t+1} =\beta ^{t+1}\cdot \frac{(1+r)^{t+1}}{1-\sigma} \cdot (1-\sigma) c_{t+1}^{1-\sigma} \quad (2) $

Dividing (2) by (1) gives $\frac{\lambda _{t+1} }{\lambda _{t}}=\beta \cdot (1+r) \frac{c_{t+1}^{1-\sigma}}{c_{t}^{1-\sigma}}$

It is the same as $\frac{\lambda _{t+1} }{\lambda _{t}}=\beta \cdot (1+r) \frac{U'(c_{t+1})}{U'(c_{t})}$

My relation is a little bit different from yours.