Is MRS supposed to have a minus sign or not? What's there to say about convexity of preferences? If we have a utility function $U(x_1, x_2) = x_1(x_2+1)^2$ of some consumer, then

$$MRS_{x_1, x_2} = \frac{\color{red}{-}(x_2+1)}{2x_1}$$



Some books have a $\color{red}{-}$. Others and Wiki don't have a minus. What exactly does this mean?



Bundle/basket/portfolio(is that only for finance?)/whatever $(3,5)$ and has a utility of $108$.

Bundle/basket/portfolio(is that only for finance?)/whatever $(27,d)$ has the same utility of $108$ for $d=1$

The MRS's are as follows:

$$MRS_{x_1, x_2}(3,5) = -1$$

$$MRS_{x_1, x_2}(27,1) = -1/27$$


  Question: What can you say about the convexity of the preferences of said consumer?


I don't quite know what to say here. The two bundles lie on the same indifference curve ($U = 108$). As we move to the right of the curve, the slope becomes less steep, as expected...I think?

This says MRS supposed to decrease if we move down a standard convex-shaped curve. So this curve is not convex? Or not standard?
The Marginal Rate of Substitution is not just the "ratio of the partial derivatives": it represents the slope of an indifference curve. In order to obtain it, you must guarantee that you remain on the same indifference curve. How do we do that? One way is by taking the total differential of the utility function and requiring this total differential to be equal to zero:

$$dU(x_1,x_2) = 0 \implies \frac {\partial U(x_1, x_2)}{\partial x_1} \cdot dx_1 + \frac {\partial U(x_1, x_2)}{\partial x_2} \cdot dx_2 =0$$

and rearranging (and putting in this case $x_2$ on the vertical axis)

$$\implies \frac {dx_2}{dx_1} = -\frac {\partial U(x_1, x_2)/\partial x_1}{\partial U(x_1, x_2)/\partial x_2} = MRS_{x_1,x_2}$$

If we look at two goods and (not say, one good and one bad), and we have non-satiation, then the partial derivatives are strictly positive. So the minus sign is the one that tells you that the slope of the indifference curve should be algebraically negative, so a "downward slopping" curve. And this is  an argument in favor of keeping it. But indeed sometimes, people omit the minus considering the negativeness of the slope to be "understood". But this can create confusion.

But hey, the symbol $dx_2/dx_1$ can also be read as "the derivative of $x_2$ w.r.t. $x_1$", which is the first derivative of the indifference curve equation. Imagine now a convex to the origin indifference curve. As you increase the value of the variable measured on the horizontal axis ($x_1$ in our case), a convex indifference curve becomes flatter. This means that in absolute terms the slope becomes smaller. But in algebraic terms it becomes larger, because its value moves from a negative number, closer to zero.

I prefer to keep signs visible at all times (many short-lived scientific discoveries have sent their discoverer into depression five minutes later, when (s)he realized that it was just a sign mistake).

So, if I want to determine whether the indifference curve is convex, I want to see the MRS (sign included) increase in value, i.e. become less negative, and so also decrease in absolute value. So we see that "MRS decreases as we move down and to the right in a convex indifference curve" uses "decrease" in the absolute value sense. 

And the confusion may get worse.

In any case, one should calculate

$$\frac {d^2x_2}{dx_1^2}$$

But with, or without the minus sign?  

Assume that, like me (and unlike many others) you want to keep the minus sign in front of the $MRS$. In that case, you want to see the MRS increase. For this to happen, the derivative of the MRS must be positive. This in my eyes is the 2nd argument in favor of including the minus sign, because we know that positive 2nd derivative means convex function (referring here to the indifference curve equation). 

So with the minus sign in front when calculating the derivative, positive $MRS$ derivative $\implies$ $MRS$ increase in algebraic value $\implies$ the slope of the $MRS$ gets larger in algebraic value, $\implies$ it gets smaller in absolute value, $\implies$ the curve gets flatter, and so it is convex indeed.
I leave the chain for the case of ignoring the minus sign as a for anyone interested. You will certainly get "negative 2nd derivative $\implies$ convex indifference curve", which will tend to mess with pre-existing knowledge in your brain.  

And check this post.