Optimal consumption policy I start with an initial capital C and at the beginning of day $n=1,...,N$ I observe the random variable $X_n$, where $\mathbb E X_n=\mu_n$. The $X_n$ are independent. I also choose $c_n$ on day $n$, the amount of my remaining capital to maximise $\mathbb E[\sum_{n=1}^{N}X_n\log(c_n)]$ subject to $\sum_{n=1}^{N}c_n=C$.
Now I want to find out the optimal consumption policy, i.e concrete values for the $c_n$. My first idea was to use Lagrange Multiplier but all I get is $-\mu_n/c_n=\lambda$ I do not see how this helps me to find the consumption policy, may you have some idea. 
 A: First of all, it doesn't make sense to consume the same amount every time period (as discussed above), at least if I am interpreting the problem correctly.  If you are sitting at time $t$ and $X_t$ is relatively large compared to $\mu_s$ for $s > t$, you should consume a lot now, since you have a high marginal utility of consumption!
One way to solve this is the dynamic programming approach, which is very popular in economics.  To do this, first define the value function, which you are trying to maximize:
$V^t(X_t, K_t) = E_t\sum_{k=0}^N \left(X_{t+k} \ln C_{t+k}\right)$
The value function is a function of the "state" you are in: $X_t$ is the current multiplier on the log function, and $K_t$ is the amount of money you have left.  So this function just tells you how well-off you are given your current state.
The reason I put a superscript on the $V$ is that the value function will not generally be the same in every time period.  This only happens in "stationary" problems (such as problems with an infinite horizon).
Now work backwards.  We know that at the final time,
$V^N(X_N, K_N) = X_N \ln K_N$
since you eat everything that is left.
At time $N-1$, the value function can be defined recursively as the sum of a "flow payoff" for time $N-1$ plus a "continuation value" for the future period $N$:
$V^{N-1}(X_{N-1}, K_{N-1}) = \max_{K_N} \left[X_{N-1} \ln \left(K_{N-1} - K_N\right) + E_{N-1}\left[V^N(X_N, K_N)\right]\right]$
This is the so-called "Bellman equation."  Note that consumption is the expression $K_{N-1} - K_N$ inside the log function.  
We can substitute in $V^N(X_N, K_N)$ as well to get
$V^{N-1}(X_{N-1}, K_{N-1}) = \max_{K_N} \left[X_{N-1} \ln \left(K_{N-1} - K_N\right) + E_{N-1}\left[X_N \ln K_N\right]\right]$
Now solve the maximization by setting the derivative with respect to $K_N$ equal to zero:
$0 = -\frac{X_{N-1}}{K_{N-1} - K_N} + E_{N-1}\left[\frac{X_N}{K_N}\right]$
We can now substitute in $\mu_N$ to get
$X_{N-1} = (K_{N-1} - K_{N}) \mu_N/K_N$
or
$K_N \left(1+\frac{\mu_N}{X_{N-1}}\right)= K_{N-1}\frac{\mu_N}{X_{N-1}}$
Multiply both sides by $X_{N-1}$ and simplify to get
$K_N = K_{N-1}\frac{\mu_N}{X_{N-1}+\mu_N}$
From which we can get the optimal consumption policy (remember that $C_{t-1} = K_{t-1} - K_{t}$).  You can substitute these values back into the Bellman equation to get the expression for the maximized value of the objective, as you asked for above.
Yes, it will be a bit annoying to continue working backwards step-by-step, but my guess is that a clear pattern will emerge as these problems are very recursive.
