The second derivative of $\log\left(\sum\limits_{i=1}^ne^{x_i}\right)$ seems negative, but I have to prove the function is convex For the second derivative I got $$\frac{\partial^2}{\partial x_k x_j}\log \left(\sum_{i=1}^{n} e^{x_i}\right)=-\frac{e^{x_k}e^{x_j}}{\left(\sum_{i=1}^{n} e^{x_i}\right)^2},$$ where $j \neq k$, and $1 \le j,k \le n$. 
This Hessian is negative, which can't allow $\log(\sum_{i=1}^{n}{e^{x_i}})$ to be convex, but I am asked to show that $\log(\sum_{i=1}^{n}{e^{x_i}})$ is convex. Obviously something is wrong here, so unless the second derivative I showed is actually positive, then I must have just gotten the wrong Hessian. 
Someone helped me calculate that:
$$\frac{\partial^2}{\partial x_k^2}\log \left(\sum_{i=1}^{n} e^{x_i}\right)=\frac{e^{x_k}\left(\sum_{i=1}^{n} e^{x_i}-e^{x_k}\right)}{\left(\sum_{i=1}^{n} e^{x_i}\right)^2},$$

There is nothing wrong. What you have only shown is not that the Hessian matrix $H$ is negative definite, but merely that the off-diagonal entries of $H$ are negative. The matrix $\pmatrix{1&-1\\ -1&1}$, for instance, has negative off-diagonal entries, but the matrix itself is positive semidefinite.
In your case, it only takes a little more work to show that the $i$-th diagonal entry of the Hessian matrix $H$ is given by
$$
\frac{e^{x_i}}{\sum_{i=1}^ne^{x_i}}-\left(\frac{e^{x_i}}{\sum_{i=1}^ne^{x_i}}\right)^2.
$$
Therefore $H$ is a (weakly) diagonally dominant matrix with a nonnegative diagonal, meaning that its eigenvalues have nonnegative real parts (Gershgorin disc theorem). As $H$ is also real symmetric, it has real eigenvalues. Hence all its eigenvalues of $H$ are nonnegative, i.e. $H$ is positive semidefinite.
