Does Littlewood's bound on $\zeta(1+it)$ extend to all the partial sums? Littlewood established that $2e^{\gamma} \geq \limsup_{t \to \infty} |\zeta(1+it)| / \log{\log{t}} \geq e^{\gamma}$, the lower bound unconditionally and the upper bound on RH. It now seems to be generally believed that the lower bound represents the truth, and even, in the most optimistic form, that quite possibly there is an absolute constant $C$ such that $|\zeta(1+it)| \leq e^{\gamma}(\log{\log{t}} + \log{\log{\log{t}}}+C)$ for all $t > 10$.
Suppose in these questions that we replace $|\zeta(1+it)|$ by 
$$
B(t) := \sup_k \Big| \sum_{n=1}^k n^{-1-it} \Big|, \quad \textrm{ resp. } \widetilde{B}(t):= \sup_{m,k} \Big| \sum_{n=m}^k n^{-1-it} \Big|,
$$
the largest partial sums, respectively the largest subsum over all intervals.
Does Littlewood's RH result extend to $B$ or even $\widetilde{B}$? That is: should the exponential sums bound $\widetilde{B}(t) = O(\log{\log{t}})$ be possible to prove on RH, or what kind of bound is available in this uniformity? Should it be reasonable to expect the strongest possible bound $\widetilde{B}(t) \leq e^{\gamma}(\log{\log{t}} + \log{\log{\log{t}}}+C)$, for all $t$?
There are evident versions of this question for $L(1,\chi)$, $\frac{\zeta'}{\zeta}(1+it)$ and $\frac{L'}{L}(1,\chi)$. In the last of these, the strongest bound seems to tie well with the belief that the smallest quadratic non-residue mod $q$ is $\ll \log{q}\log{\log{q}}$ -- which it certainly implies, at least when $q$ is prime. 
 A: The short answer is yes, and this is treated explicitly for characters in the work of Granville and Soundararajan (the paper appeared in J. Amer. Math. Soc.).  Their Theorem 2 gives that on GRH for $x\le q$ and a primitive character $\chi \pmod q$ one has 
$$ 
\Big| \sum_{n\le x} \chi(n) \Big| \ll \Psi(x, (\log q)^2 (\log \log q)^{20}), 
$$ 
where $\Psi(x,y)$ denotes the number of integers up to $x$ composed only of prime factors below $y$.  From this and partial summation, it follows at once that for $x\le q$
$$ 
\Big| \sum_{n\le x} \frac{\chi(n)}{n} \Big| \ll \prod_{p\le (\log q)^2(\log \log q)^{20}} \Big(1- \frac{1}{p}\Big)^{-1} \ll \log \log q. 
$$ 
In fact their Theorem 2 is a bit more precise, and one should be able to get the Littlewood type constant also.  This result will also extend to the $t$-aspect with minor changes, and establish the bound that you want.  
Alternatively you could look at another paper of Granville and Soundararajan (also in JAMS, and also in the case of characters); the argument given there in Section 5 would give a slightly simpler approach to such results.  
Here's a quick sketch proof based on the second approach.  Let $|t|$ be large, and assume that $x\le |t|$ (for larger $x$ the partial sum simply approximates $\zeta(1+it)$).  Let $y$ be a parameter to be chosen, and note that $n\le x$ is either $y$-smooth, or may be written as $n=mp$ where $p\ge y$ is the largest prime factor of $n$ (so that $m$ is $p$-smooth).  Thus, with $P(m)$ denoting the largest prime factor of $m$,
$$ 
\sum_{n\le x} \frac{1}{n^{1+it}} = \sum_{\substack{n\le x\\ p|n \implies p\le y}} \frac{1}{n^{1+it}} + \sum_{m\le x/y} \frac{1}{m^{1+it}} \sum_{\max(y, P(m)) \le p \le x/m} \frac{1}{p^{1+it}}.
$$ 
The first term in the RHS is simply bounded by $\prod_{p\le y} (1-1/p)^{-1}$.  As for the second term, RH can be used (usual contour shift argument) to show that the sum over $p$ there is 
$$ 
\ll \frac{(\log |t|)}{\sqrt{y}},  
$$ 
and bounding the sum over $m$ trivially, the second term is 
$$ 
\ll \frac{(\log |t|) \log x}{\sqrt{y}} \ll \frac{(\log |t|)^2}{\sqrt{y}}. 
$$ 
Now choosing $y=(\log |t|)^4$ gives a bound of $\ll \log \log |t|$ for your partial sums.  With more care the product may be truncated at $y=(\log |t|)^{2+\epsilon}$.  One would expect that truncation at $y=(\log |t|)^{1+\epsilon}$ (or even $y=C \log |t| \log \log |t|$) is the truth.
