Let's consider the number \begin{equation} \xi_1 = \sum_{n=1}^\infty \frac{(-1)^{n^\textrm{th}\textrm{ digit of $\pi$ in base $2$}}}{n} = -1 -\frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} - \frac{1}{13} - \dots \end{equation} where I ignored the dot separation of integer part $11$ with the rest of $\pi$ binary representation. I'm not sure a such number exists: clearly the series is not absolutely convergent, and I don't know how to check convergence of a such "chaotic" (so to speak) sign changing series. I did some numerical explorations to see if they can support the hypothesis that the series converges, exploiting the formula $a_n = \textrm{floor}\left( \pi \cdot {{2}^{n-3}}\right) -2 \textrm{floor}\left( \pi \cdot {{2}^{n-4}}\right)$ found in online encyclopedia of integer sequences: if $n>1$ it gives the sequence of digit of $\pi$ written in base $2$ (leaving away the dot separation of integer part, that is given by the first two digit): \begin{equation} 1100100100001111110110\dots \end{equation} So \begin{equation} \xi_1 = \sum_{n=2}^\infty \frac{(-1)^{\textrm{floor}\left( \pi \cdot {{2}^{n-3}}\right) }}{n-1} \end{equation} where I left away the useless unitary denominator $(-1)^{2\cdot\textrm{integer}}$. Now, if $x\lesssim 50$ the function $f_1(x)=\sum_{n=2}^x \frac{(-1)^{\textrm{floor}\left( \pi \cdot {{2}^{n-3}}\right)}}{n-1}$ doesn't look so regular:
but if we go on with the sum, things changes: a regular trend is graphically found:
Unluckily, this trend is very vague is it not sufficient not even to try to guess the limit ($e$? $+\infty$? $\dots$?). My scrap worked more than 3 minutes to plot this, and goes out of memory if with $x>1025$ (we have $f(1025) \approx 2.24806174858$), but surely a more powerful computer can explore deeper this function.
Similarly, what about this series? \begin{equation} \xi_2 = \sum_{n=1}^\infty \frac{{n^{\textrm{th}}\textrm{ digit of $\pi$ in base $2$}}}{n} = 1 + \frac{1}{2} + \frac{1}{5} + \frac{1}{8} + \frac{1}{13} + \frac{1}{14} + \dots \end{equation} Working as done before, we have $\xi_2 = \sum_{n=2}^\infty \frac{ \textrm{floor}\left( \pi \cdot {{2}^{n-3}}\right) -2 \textrm{floor}\left( \pi \cdot {{2}^{n-4}}\right) }{n-1}$. This time, the sequence of partial sums is non decreasing. Let's consider the function \begin{equation} f_2(x) = \sum_{n=2}^x \frac{ \textrm{floor}\left( \pi \cdot {{2}^{n-3}}\right) -2 \textrm{floor}\left( \pi \cdot {{2}^{n-4}}\right) }{n-1} \end{equation} If $x=10$ this gives $1 + \frac{1}{2} + \frac{1}{5} + \frac{1}{8} = \frac{73}{40}=1.825$. If $x=20$ it gives $\frac{2469269}{1113840}\approx 2.2168974$. When $x=84$ the value is \begin{equation} \frac{433233218232720513501779}{153615744795583006700400} \approx 2.820239675361568 \end{equation} and for $x > 84$ my computer cannot find the fraction and any change in the approximated value. This suggests that the limiting value (never reached: otherwise $\pi$ would be rational) could be a little bigger than 2.82.