Is Weighted Averages the Best Method to Aggregate Information?
A weighted average might or might not be the best method.  This will depend on (a) the nature of the algorithms, and (b) what objective function you are trying to minimize.

Let me give you a sample set of assumptions.  Suppose we treat the predictions from the recommendation algorithms as independent random variables (e.g., because each recommendation algorithm was trained independently).  Also,  suppose we wish to minimize the variance of the final prediction.  In other words, our loss function is the $L_2$ loss: if our prediction was $Y$, and the correct value was $\tilde{Y}$, then the loss (penalty) is $(Y-\tilde{Y})^2$.

Then in this case we can find the optimal set of weights.  In particular, in this case, if we restrict ourselves to linear combinations, the optimal set of weights is inversely proportional to the variance of each estimator.  Suppose $X_1,\dots,X_n$ are the predictions from our $n$ recommendation algorithms.  Suppose that we have measured the accuracy of each recommendation algorithm by measuring its $L_2$ loss on a validation set; in other words, we know the variance $\text{Var}(X_i)$ for each $i$.  Here $\text{Var}(X_i)$ is a measure of the accuracy of the $i$th recommendation algorithm: the larger the variance, the larger the expected $L_2$ loss, and thus the less accurate the prediction is.  Then the optimal linear weighting is to select a final prediction $Y$ by

$$Y = c_1 X_1 + \dots + c_n X_n,$$

where we choose the weights

$$c_i = {a_i \over a_1 + \dots + a_n}$$

where

$$a_i = {1 \over \text{Var}(X_i)}.$$

Is a linear combination optimal (as opposed to other, nonlinear methods of combining the predictions)?  In this situation, if we additionally know that the errors made by each recommendation algorithm have a Gaussian distribution, then yes, a linear combination is optimal.  If the errors have a different distribution, then some other combination might be better.

As you can see, this will depend on the loss function we select, and on other assumptions.  That said, in practice, a linear combination (i.e., a weighted average) is often a reasonable choice.

There's lots written about this sort of thing in the statistics literature.  For instance, the above discussion is just a restatement of the fact that the average is the optimal estimator of the mean of a Gaussian distribution (i.e., it minimizes the mean squared error).  See, e.g., https://en.wikipedia.org/wiki/Estimator, https://en.wikipedia.org/wiki/Mean_squared_error, https://stats.stackexchange.com/q/81571/2921, https://stats.stackexchange.com/q/97765/2921, https://stats.stackexchange.com/q/48864/2921, https://math.stackexchange.com/q/9032/14578.