Let $X$ be a Polish space and let $(\mu\_i)\_{i=1}^{\infty}$ be a sequence of probability measures in the Wasserstein space $\mathcal{P}(X)$ on $X$. Let $(\beta\_i)\_{i=1}^{\infty}$ be a summable sequence in $(0,\infty)$. For every positive integer $k$, define the probability measures
$$
\nu\_k = (\sum\_{1\leq i\leq k}\beta\_i)^{-1} \sum\_{1\leq i\leq k} \beta\_i \mu\_i
$$
and define the probability measure
$$
\nu\_{\infty} = (\sum\_{i=1}^{\infty}\beta\_i)^{-1} \sum\_{i=1}^{\infty} \beta\_i \mu\_i.
$$


Is it true/clear that
$
\lim\limits\_{k\to\infty} \mathcal{W}(\nu\_k,\nu\_{\infty}) = 0,
$ quantitatively (here $\mathcal{W}$ denotes the Wasserstein distance on $\mathcal{P}(X)$).



Wasserstein convergence of "series expansion'' of probability measure
It is true and clear if the metric space $X$ has a finite diameter, but false in general: Take $\beta\_i=2^{-i}$ and $\mu\_i$ the point mass at $3^i$.


Details: In the case $D=$diam$(X)<\infty$, write $s\_k=\sum\_{i \le k} \beta\_i$
and $s=\sum\_{i<\infty} \beta\_i$. Then $$\nu\_\infty=\frac{s\_k}{s} \nu\_k+\frac{s-s\_k}{s} \gamma\_k$$
for some probability measure $$\gamma\_k=\frac{\sum\_{i > k} \beta\_i \mu\_i}{s-s\_k} \,.$$
Therefore $$\nu-\nu\_k=\frac{s-s\_k}{s} (\gamma\_k-\nu\_k)$$
whence $$\mathcal{W}(\nu\_k,\nu\_{\infty}) \le \frac{s-s\_k}{s} D$$
because the diameter of the Wasserstein metric on $\mathcal{P}(X)$ is $D$.


