Monotonicity of Riemannian Logarithm Let $(X,g)$ be a complete, connected, Riemannian manifold of non-positive curvature.  Let $(z_n)_{n=1}^{\infty}$ be a sequence in $X$ converging to some $z \in X$ such that
$$
d_g(z,z_{n+1})\leq d_g(z,z_n),
$$
for every $n\geq 1$; where $d_g$ is the intrinsic metric induced by the riemannian metric $g$.  By the Cartan-Hadamard theorem, we know that in this case each $\exp^{-1}_x$ is globally well-defined on $X$.
Is it then true, that the sequence of functions
$$
\begin{aligned}
F_n:X^2 &\rightarrow [0,\infty)\\
(x,y) & \mapsto \|\exp_{z_n}^{-1}(y)-\exp_{z_n}^{-1}(x)\|,
\end{aligned}
$$
converges to $(x,y)\mapsto \|\exp_{z}^{-1}(y)-\exp_{z}^{-1}(x)\|$ uniformly and monotonically (ie $F_{n+1}(x,y)\leq F_{n}(x,y)$ for each $x,y \in X$?)
 A: 
This is not true. Assume $X$ is the hyperbolic plane (with constant curvature $-1$). Fix any $n$, let $w=d(z_n, z)$. Extend the geodesic from $z_n$ to $z$ by distance $b$ and reach a point $x$; and at $z$ draw a geodesic of length $b$ that is perpendicular to the first geodesic, reach a point $y$. Here $b$ is a large number depending on $w$. Write  $a=d(z_n, y)$. So
$$
|\exp_z^{-1}(y)-\exp_z^{-1}(x)|=\sqrt 2 b. 
$$
Now by the hyperbolic law of cosine,
$$
\cosh a=\cosh b \cdot \cosh w.
$$
Now $a, b$ are both big, $\cosh b=\frac{e^{b}}{2}(1+e^{-2b})$, so $\log \cosh b=b-\log 2+O(e^{-2b})$. $w$ is small, so $\log\cosh w=\log(1+\frac{w^2}{2}+...)
=\frac{w^2}{2}+O(w^4)$. So
$$
a-\log 2+O(e^{-2b})=b-\log 2+O(e^{-2b})+\frac {w^2}{2}+O(w^4),
$$
i.e. $a=b+\frac{w^2}{2}+O(e^{-2b})+O(w^4)$. We can take $b$ so big that we have $e^{-2b}\ll w^4$, so
$$
a=b+\frac{w^2}{2}+O(w^4).
$$
Apply hyperbolic law of cosine to the same thin triangle $z_nzy$ we get
$$
\cos\theta=\frac{\cosh a\cosh w-\cosh b}{\sinh a\sinh w}
=\frac{\cosh b\cosh w\cosh w-\cosh b}{\sinh a\sinh w}
=\frac{\cosh b\sinh^2 w}{\sinh a\sinh w}=\frac{\cosh b\sinh w}{\sinh a}. 
$$
From $a=b+\frac{w^2}{2}+O(w^4)$ and $b$ is sufficiently big, this implies
$$
\cos\theta=w+O(w^2).
$$
This is the key estimate, saying that $\theta$ differs from the right angle $\pi/2$ by $O(w)$, much bigger than the Euclidean case $O(w/a)$.
So, finally,
$$
\begin{aligned}
|\exp_{z_n}^{-1}(y)-\exp_{z_n}^{-1}(x)|=&\sqrt{(b+w)^2+a^2-2(b+w)a\cos\theta}\\
=&\sqrt{b^2+w^2+2bw+b^2+b\cdot O(w^2)-2(b+w)(b+O(w^2))(w+O(w^2))}\\
=& \sqrt{2b^2+w^2+2bw+b\cdot O(w^2)-2b^2w+b^2O(w^2)}\\
=& \sqrt{2b^2+2bw-2b^2w+b^2O(w^2)}\\
=& \sqrt 2 b\sqrt{1+\frac wb-w+O(w^2)}\\
=& \sqrt 2 b\Big(1-\frac w2 +\frac {w}{2b} +O(w^2)\Big);
\end{aligned}
$$
this differs from the limit $\sqrt 2 b$ by at least $\frac{\sqrt 2 b w}{3}>1$, if $b$
is sufficiently big.
So the converge is not uniform. And there is no reason to think monotone (try travel backwards in a direction from $z$ to $z_n$...)
