Does this sum converge or diverge? Does the infinite sum $\large{\sum_{n=1}^\infty \frac{1}{n^{x_{\small{n}}}}}$ converge if $x_n$ is a random variable (generated within each term) that takes values between $0$ and $2$ with equal probability converge or diverge? I have a suspicion that it diverges, but I don't know how to prove it.
 A: If $P(x_n=0)=c$ for every $n$ with $c\gt0$ and the sequence $(x_n)$ is independent, then the random set $Z=\{n\geqslant1\mid x_n=0\}$ is almost surely infinite and
$$
\sum_{n\geqslant1}\frac1{n^{x_n}}\geqslant|Z|,
$$
hence the series on the LHS diverges almost surely.
The same result holds, assuming only independence and that $P(x_n\leqslant1)\geqslant c$ for every $n$, for some $c\gt0$, but the proof is slightly more elaborate. 
To see how the proof works, assume as in the question that $(x_n)$ is i.i.d. and uniformly distributed on the interval $(0,2)$ and consider, for every $n$, the random set $$L_n=\{k\mid 2^n\leqslant k\lt2^{n+1}, x_k\leqslant1\},$$ of size $\ell_n=|L_n|$, and the part $S_n$ of the series restricted to $L_n$, that is, 
$$
S_n=\sum_{k=2^n}^{2^{n+1}-1}\frac1{k^{x_k}}.
$$
For every $k$ in $L_n$, $k^{x_k}\leqslant k\leqslant2^{n+1}$ hence
$$
S_n\geqslant\sum_{k\in L_n}\frac1{k}\geqslant\frac{\ell_n}{2^{n+1}}.
$$
For every $k$, let $U_k=\mathbf 1_{x_k\leqslant1}$, then $(U_k)$ is i.i.d. Bernoulli with parameter $\frac12$ and 
$$\ell_n=\sum\limits_{k=2^n}^{2^{n+1}-1}U_k
$$ is binomial $(2^n,\frac12)$ with mean $E(\ell_n)=2^n\cdot\frac12$ and variance $\sigma^2(\ell_n)=2^n\cdot\frac14$ hence 
$$
[\ell_n\leqslant2^{n-2}]\subset[|\ell_n-E(\ell_n)|\geqslant2^{n-2}],
$$
and Bienaymé-Chebychev inequality yields
$$
P(\ell_n\leqslant2^{n-2})\leqslant\frac{\sigma^2(\ell_n)}{(2^{n-2})^2}=\frac4{2^n}.
$$
The RHS is summable hence Borel-Cantelli lemma implies that, almost surely, for every $n$ large enough,  $\ell_n\gt2^{n-2}$. When $\ell_n\gt2^{n-2}$,
$$
\sum_{k=2^n}^{2^{n+1}-1}\frac1{k^{x_k}}\geqslant\frac18.
$$
The RHS does not converge to $0$. This implies the almost sure divergence of the full series
$$
\sum_{k=1}^{\infty}\frac1{k^{x_k}}.
$$
The proof when one assumes only that $(x_n)$ is independent and that $P(x_n\leqslant1)\geqslant c$ for every $n$, for some $c\gt0$, is quite similar.
A: Since the series in question is based on random values, it does not make sense to ask if it converges; rather one should ask about the probability that it will converge to a finite value.
Define the Boolean function
$$
b(x) =
\begin{cases}
0, \text{if $x \geq 1$} \\
1, \text{if $x \lt 1$}
\end{cases}
$$
Now set $b_n = b(x_n)$.  Clearly, $0 \lt 1/n^{x_n}$, and if $x_n < 1$, then $1/n < 1/n^{x_n}$.  Thus it must always be the case that  $$b_n / n \lt 1/n^{x_n}$$
Likewise, the sum:
$$
\sum_{n=1}^\infty {b_n\over n} \lt \sum_{n=1}^\infty {1\over n^{x_n}}
$$
Since each $x_n$ is randomly (evenly) distributed over $0 \lt x_n \lt 2$, the $b_n$ are distrbuted evenly as $0$ or $1$.
(Edited here to be more formal)
Now, define a new series $c_m$ with $c_0 = b_1$ and
$$
c_m = \sum_{n=2^{m-1}+1}^{2^{m}} \frac{b_n}{n}  (m \geq 1)
$$
This is just separating the $b_n/n$ terms into groups.  Each $c_m$ is a sum of $2^{m-1}$ terms (except for $c_0$, which has $1$ term, not $\frac12$).  Since every $b_n/n$ is part of exactly one $c_m$, we must have:
$$
 \sum_{m=0}^\infty c_m = \sum_{n=1}^\infty {b_n\over n}
$$
Each $c_m$ is a sum of $2^{m-1}$ terms, and each term has a denominator less than $2^{m}$.  If all the $b_n$ are $1$, then $c_m > 1/2$.  On average, half the $b_n$ are $1$, so the average value for each $c_m > 1/4$.    Specifically, $c_m < 1/4$ only if more than half of the related $b_n$ are $0$.  Since each $b_n$ is independently $0$ or $1$ with $50\%$ probability, the probability that more than half of them are $0$ is clearly less than $1/2$.
The sum
$$
\sum_{m=1}^\infty c_m
$$
is a sum of an infinite number of terms, each of which is $>1/4$ with $>50\%$ probability.  This series clearly diverges. More formally, it should be straightforward to demonstrate:
$$
\forall (T>0) \forall (\epsilon>0): \exists (N) \mid 
P\left(\sum_{m=0}^N c_m < T\right) < \epsilon
$$
That is one definition of "divergent" for a probabilistic series.
From above:
$$
\sum_{n=1}^\infty {1\over n^{x_n}} > \sum_{n=1}^\infty {b_n\over n} = \sum_{m=0}^\infty c_m
$$
The last term diverges, so the others must also.
A: NOTE: At the time this post was written, the problem was $\sum_n \frac{1}{n^x}$ - the asker has revised their problem statement. 
With probability $\frac{1}{2}$, the series diverges (it is a $p$-series with $p\leq 1$ for $x \in [0,1]$) and with probability $\frac{1}{2}$ the series converges (it is a $p$ series with $p>1$ for $x \in (1,2]$). 
Note the random variable $x$ is fixed over the sum, so it just becomes a calc 1 convergence problem for each value $x$ takes. 
A: I don't know if I exactly have a proof, but here's a thought.  The infinite series of reciprocals of the prime numbers diverges.  How likely is it that $ \ x_n \ > \  1 \ $ "often enough" to produce a series with terms that can bring the series to convergence despite that?  That is, can there be a high enough "density" of terms that make the series convergent against the sum of terms that would cause divergence?  Perhaps there is an argument something like comparing $ \ \sum_{n=1}^{\infty} \ \frac{1}{n^{x_n}} \  $ to $ \ \sum_{n=1}^{\infty} \ \frac{1}{p_n} \  $ , which has a lower "density" of terms than, say, the harmonic series.  (I suspect the probability of having a convergent series is essentially zero.)
