Are there any series whose convergence is unknown? Are there any infinite series about which we don't know whether it converges or not? Or are the convergence tests exhaustive, so that in the hands of a competent mathematician any series will eventually be shown to converge or diverge?
EDIT: People were kind enough to point out that without imposing restrictions on the terms it's trivial to find such "open problem" sequences. So, to clarify, what I had in mind were sequences whose terms are composed of "simple" functions, the kind you would find in an introductory calculus text: exponential, factorial, etc.
 A: Let
$$h(x) = \sum_{k=1}^\infty \frac{\{2^kx\}-\frac{1}{2}}{k},$$
where the curly brackets represent the fractional part function, and $x\in [0,1]$. For almost all numbers $x$, it is not know whether the series converge or not. It is believed to converge for almost every number, yet it is very hard to come up with a single non-rational number, not one that is artificially manufactured, such that the convergence status is known. If for instance you can prove that it converges for $x = \frac{\pi}{4}$ (everyone strongly believes that it does, and it is backed by empirical evidence), you would instantly become very famous in the math community. Chances are that it is impossible to prove or disprove convergence for $x=\pi,e,\log 2$ and most other math constants. This question was raised in section 4.3.(a) in this article, where you can find more details about it.
It would be interesting to study the wildly erratic behavior of this function, which is not only discontinuous everywhere, but admits a dense set of singularities (where it does not converge.)
A: It is unknown whether the series:
$$\sum_n \frac{(-1)^n n}{p_n}$$
converges. Here, $p_n$ is the $n$-th prime number. This problem is posed in Guy's book on unsolved problems in number theory and I am pretty sure that it originated from Erdős.  
A: From Wikipedia:
The statement that the equation:
$$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$$
is valid for every $s$ with real part greater than 1/2, with the sum on the right hand side converging, is equivalent to the Riemann hypothesis.
This in turn can be rewritten as:
$$\displaystyle \frac{1}{\lim\limits_{k\to \infty } \, \left(\sum\limits_{a=1}^{k} \frac{1}{a^s}+\frac{1}{(s-1) k^{s-1}}\right)} =$$
$$\displaystyle \lim_{k \rightarrow \infty} \left( \underbrace{1 - \sum_{2 \leq a \leq k} \frac{1}{a^{s}} + \underset{ab \leq k}{\sum_{a \geq 2} \sum_{b \geq 2}} \frac{1}{(ab)^{s}} - \underset{abc \leq k}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} \frac{1}{(abc)^{s}} + \underset{abcd \leq k}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} \frac{1}{(abcd)^{s}} - \cdots}_{\text{number of alternating sums} > \frac{\log(k)}{\log(2)}} \right)$$
where:
$\Re(s)>0:\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\zeta(s)=\lim\limits_{k\to \infty } \, \left(\sum\limits_{n=1}^{k} \frac{1}{n^s}+\frac{1}{(s-1) k^{s-1}}\right)$
A: As kind of a joke answer, but technically correct, and motivated by Chandru's deleted reply,
$$\sum_{n=0}^\infty \sin(2\pi n!\,x)$$
where $x$ is the Euler-Mascheroni constant, or almost any other number whose rationality has not been settled. (If $x$ is rational, the series converges. The implication does not go the other way.)
A: Reuns answer would be:
$$\displaystyle \sum_{n=2}^{\infty} \left( \underbrace{-\frac{1}{\sqrt{n} \log^{3+\epsilon}(n)}+\underset{a = n}{\sum_{a \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} - \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} + \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} - \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} + \cdots}_{\text{number of alternating sums} > \frac{\log(n)}{\log(2)}} \right)$$
Mathematica: https://pastebin.com/gxAE6ZgY
A: The Riemann hypothesis is that $\sum_{n=2}^\infty \frac{\Lambda(n)-1}{n^{1/2}\log^{3+\epsilon} n}$ converges for any $\epsilon > 0$ (see here for a discussion of that $\epsilon$).
A: It is unknown whether the Flint Hills series
$$
\sum_{n=1}^\infty\frac{1}{n^3\sin^2n}
$$
converges or not. The difficulty here is that convergence depends on the term $n\sin n$ not being too small, which in turn depends on how well $\pi$ can be approximated by rational numbers. It is possible that, if $\pi$ can be approximated `too well' by rationals, then this will diverge. See this MathOverflow question for a discussion of this particular series.
Another even simpler example of a sequence (no summation) for which it is not known whether it converges or not is
$$
x_n=\frac{1}{n^2\sin n}.
$$
We would expect this to tend to zero, but the proof is beyond what is currently known. Suppose that there were only finitely many rational numbers $p/q$ with $\vert p/q-\pi\vert\le q^{-3+\epsilon}$ (for any $\epsilon > 0$), then $x_n$ would tend to zero at rate $O(n^{-\epsilon})$. If, on the other hand, there were infinitely many rationals satisfying $\vert p/q-\pi\vert\le q^{-3-\epsilon}$, then infinitely many $x_n$ would be of order at least $n^\epsilon$, so it diverges. This can be expressed in terms of the irrationality measure of $\pi$. The sequence $x_n$ converges to zero if the irrationality measure of $\pi$ is less than 3, and diverges if it is geater than 3. Currently, the best known bound for the irrationality measure is that it is no more than about $7.6063$* (see the link to the mathworld page above). It is expected that the irrationality measure of $\pi$ is 2 (it is known that all but a zero-measure set of real numbers have irrationality measure 2). Therefore, it is expected that $x_n$ tends to zero, but there is currently no proof of this.
[* The best known bound for the irrationality measure, as listed on the mathworld page, has been improved! It is now 7.10320533, according to the (not yet peer-reviewed) paper by Zeilberger and Zudlin, 2019. This is still much too small an improvement to say whether or not the sequence above converges.]
A: The Riemann hypothesis is equivalent to:
$$\lim_{n\to \infty } \, \frac{\sum\limits_{k=1}^n \lambda (k)}{n^{\frac{1}{2}+\epsilon}}=0$$
where $\lambda(k)$ is the Liouville Lambda function.
This in turn can be rewritten as:
$$\lim\limits_{n \rightarrow \infty}\frac{1}{n^{\frac{1}{2}+\epsilon}}\left(\underbrace{\underset {a^2 \leq n} {\sum_ {a\geq 1}} 1 - \underset {a^2 b \leq n} {\sum_ {a\geq 1}\sum_{b\geq 2}} 1 + \underset {a^2 bc \leq n} {\sum_ {a\geq 1}\sum_ {b\geq 2}\sum_ {c\geq 2}} 1 - \underset {a^2 bcd \leq n} {\sum_ {a\geq 1}\sum_ {b\geq 2}\sum_ {c\geq 2}\sum_ {d\geq 2}} 1 + \cdots}_{\text{number of alternating sums} > \frac{\log(n)}{\log(2)}}\right)=0$$
