Consider the following heat diffusion problem.
So, $\lim\limits_{t\to\infty} u (x, t)$ is equal to:
The solution is $\frac {L^2}3$
I only arrive to $$u(x,t)= \frac {a_0}2 + \sum_{n=1}^{\infty} a_n \cos(\frac {n\pi x} L) e^{\frac {-4\pi^2n^2t}{L^2}}$$
If you want to reason intuitively, then you could do so as follows: the Neumann BC's tell you that there is no heat loss at the boundaries, so the total amount of heat $$Q(t) = \int_0^L u(x,t) dx$$ must be conserved. You could probe this directly using integration by parts if you want, but it's clear physically. The initial amount of heat is equal to $$ \int_0^L x^2 dx = L^3 / 3.$$ As $t \to \infty$, this heat will distribute itself over a length $L$, which means that the distribution tends to a constant $u(x,t) = L^2 / 3$. (I divided by $L$ to compensate for the length of the rod).
Of course, you can also use the Fourier series. Note that each term in the summation you have is exponentially damped, to as $t \to \infty$, you are only left with the $a_0/2$ part. You can simply calculate the first term in the Fourier cosine expansion of $x^2$ to get the solution $L^2/3$.