Are there limits of which we're not able yet to find the value or not even prove non/existence? I really like working out limits, so I've been wondering:

Are there limits we're struggling to evaluate? Are there limits of which we're not succeeding in proving the existence or nonexistence?

 A: I feel like this is cheating but it is well known that
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}\\
\sum_{n=1}^{\infty} \frac{1}{n^{4}} = \frac{\pi^{4}}{90}\\
\sum_{n=1}^{\infty} \frac{1}{n^{6}} = \frac{\pi^{6}}{945}
\end{align*}
etc, but there is no known analytic form for 
$$\sum_{n=1}^{\infty}\frac{1}{n^{k}}$$
when $k$ is odd.
As another rich area that would be better left for someone else elaborate on: Newton's method. For different initial values, there is still relatively very little which is known about the behavior of the Newton iterates. (Especially considering that this is one of the most well known methods for finding roots numerically.)
A: Going similarly to @JessicaK's answer, there are plenty of other infinite series which are open. These are certainly limits, since
$$\sum_{n=1}^\infty x_n \equiv \lim_{N\to\infty}\sum_{n=1}^N x_n$$
by defintion.
A famous infinite series which we are unable to evaluate at the moment is
$$\sum_{n=1}^\infty \frac{1}{n^3\sin^2 n}\tag1$$
There's actually a MO question about this series, and no resolution is given there.
In an answer that very question, the above series is shown to be related to the "difficult limit" o
$$\vert n\sin n\vert,n\to\infty$$
Said answer ends up non-trivially relating this limit to the convergents of $\pi$. Note that the limit itself clearly goes to infinity ($\vert n\sin n\vert = \vert n\vert\vert \sin n\vert$), however investigating the behavior of the function as it tends to infinity is sort of the "next level" past just evaluating the limits.
A: You can surely cook up series whose convergence is unknown.  For example, let $d_k$ be the sequence of digits of $\pi$ (i.e., $d_0=3$, $d_1=1$, $d_2=4$, etc.).  I'm pretty sure the convergence of
$$\sum_{k=0}^\infty{(-1)^{d_k}\over k+1}=-1-{1\over2}+{1\over3}-{1\over4}-{1\over5}-{1\over6}+\cdots$$
is unknown.
