Are the solutions to $1+1/2^s+1/3^s=0$ known? For $s$ a complex number, are the solutions to this equation known?
$$1+1/2^s+1/3^s=0$$
Borwein et alia have studied the partial sums of the zeta function:
Zeros of partial sums of the Riemann zeta-function and found that the solutions have periods:
$$\frac{2 i \pi}{\log (2)}$$ and $$\frac{2 i \pi}{\log (3)}$$
but I could not understand or find what the exact solutions are.
These periods are found in the solutions of the somewhat similar:
$$1+1/2^s+1/3^s+1/6^s=0$$
which has the solutions:
$$\left\{6,c_1\in \mathbb{Z}\land \left(s=\frac{2 i \pi  c_1}{\log (2)}+\frac{i \pi }{\log (2)}\lor s=\frac{2 i \pi  c_1}{\log (3)}+\frac{i \pi }{\log (3)}\right)\right\}$$
Mathematica also knows the solutions to:
$$1/2^s+1/3^s=0$$
and:
$$\sum _{n=1}^k \frac{1}{\exp (n)^s}=0$$
but not to the first equation above.
Some related Mathematica code:
Clear[n, k, s]
Reduce[1/2^s + 1/3^s == 0, s]
Do[Print[{k, 
   Reduce[Total[Table[1/Exp[n]^s, {n, 1, k}]] == 0, s]}], {k, 1, 6}]
Clear[n, k, s]
Do[Print[{k, Reduce[Total[Divisors[k]^s] == 0, s]}], {k, 1, 6}]

 A: Without even thinking I will shoot from the hip and say no explicit closed-forms are known.
Firstly, the paper you cite does not actually say the zeros have those periods. In fact, since the numbers $\log 2$ and $\log3$ are linearly independent over $\Bbb Q$, anything with both those periods would be dense on a line. Rather, zeros of $1+2^{-s}+3^{-s}$ are the intersection points of the (complex) graphs of the two functions $1+2^{-s}$ and $-3^{-s}$ (for instance), which it is easy to check have complex periods $2\pi i/\log 2$ and $2\pi i/\log 3$ respectively.
Secondly, $a^s+b^s=0$ has two exponentials but can easily be reduced to one exponential, allowing the term $s$ to be isolated. The second sum you cite is just an unravelling of another expression with two expentials, made to appear to have more exponentials artificially using the geometric sum formula. It is also noteworthy that $1+2^{-s}+3^{-s}+6^{-s}$ factors as $(1+2^{-s})(1+3^{-s})$ and so reduces to two cases each involving only one exponential.
By contrast, perturbing a two-exponential expression by a constant is not amenable to the above type of factorization, nor is it amenable to being rescaled into just one exponential. If there were known symbolic closed-form means for solving this, or if there was a special function invented for this purpose, I'd imagine it'd be more widely known. (Much like how the Lambert W function was designed to take on situations where a single-exponential equation has been perturbed by a power of the unknown under certain conditions.)
A: It's easy to see that all zeros of $1 + 1/2^s + 1/3^s$ have $-1 \le \text{Re}(s) \le 0.787885$ (the actual upper bound is the positive solution of
$1 - 1/2^t - 1/3^t = 0$).  Here is a plot of the $70$ zeros with $0 < \text{Im}(s) < 400$, according to Maple (of course the complex conjugate of a zero is also a zero).

Although there is no periodicity, there appears to be an approximate symmetry: if $s$ is a zero then there is a zero close to (but not exactly) $480.42 i + \overline{s}$.
This comes from the fact that $480.42 \ln(3)$ and $480.42 \ln(2)$ are
quite close to multiples of $2\pi$.
A: you could deduce interesting facts on the roots of $\zeta_N(s) = \sum_{n=1}^N n^{-s}$ by computing the Dirichlet series 
$$\frac{1}{\zeta_N(s)} = \sum_{k=0}^\infty (1-\zeta_N(s))^k = 1 +\sum_{n=2}^\infty n^{-s} \sum_{\begin{array}{c}n = d_1 \ldots d_k\\ d_i \le N\end{array}} (-1)^k$$
where $\displaystyle\sum_{\begin{array}{c}n = d_1 \ldots d_k\\ d_i \le N\end{array}}$ is the sum on all the possible factor decomposition of $n$ where the order counts and each factor $1 < d_i \le N$
A: We can find 38 solutions in a row with the Mathematica 8 program:
(*Mathematica 8 start*)
Clear[nn, cc, s, x, y];
Print["k can be varied to any integer greater than or equal to 2:"]
k = 3;
cc = 2000;
s = 0;
Do[
 Do[s = (2 I \[Pi]*(Floor[kk*Exp[1] - 1]))/Log[k] + 
    N[Round[Log[1/(-Sum[1/n^s, {n, 1, k - 1}])]/Log[k]*10^80]/10^80, 
     80], {i, 1, cc}];
 Print[{kk, Sum[1/n^(s), {n, 1, k}]}], {kk, 1, 100}]
(*end*)

It fails the first time at kk=39, and it probably skips a lot of zeros in between.
I don't know why it works though, when it works.
The first few values of kk in the program above when it does not find zeros are:
39,42,46,53,60,63,66,67,70,73,74,77,78,81,84,85,87,88,91,92,94,95,97,98,99,...
The first few values of kk when the program below does find zeros, almost matches the sequence above with only 2 exceptions right at the beginning:
1,35,39,42,46,49,53,56,60,63,66,67,70,73,74,77,78,80,81,84,85,87,88,91,92,94,95,97,98,99,...
(*Mathematica 8 start*)
Clear[nn, cc, s, x, y];
Print["k can be varied to any integer greater than or equal to 2:"]
k = 3;
cc = 2000;
s = 0;
Do[Do[s = (2 I \[Pi]*(Floor[kk*Exp[1] - 2]))/Log[k] + 
    N[Round[Log[1/(-Sum[1/n^s, {n, 1, k - 1}])]/Log[k]*10^80]/10^80, 
     80], {i, 1, cc}];
 Print[{kk, Sum[1/n^(s), {n, 1, k}]}], {kk, 1, 100}]
(*end*)

But both programs are most likely skipping zeros.
