Recall that a complex manifold is Stein if it is holomorphically convex and separable. If we assume holomorphically convex alone, then there is Cartan-Remmert reduction to say how far it is from being Stein. If we assume holomorphic separation alone, I learnt from Th 9.9(b) in p41 of Demailly's lecture notes [Here](http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/estimations_l2.pdf) that:


"On a complex manifold X, if $O(X)$ locally separate points (i.e. for any $x \in X$, there is a nbhd of $U\_x$ such that for any $y \in {U\_x}-x$, there exists $f \in O(X)$ with $f(x) \neq f(y)$), then there exists a smooth nonnegative strictly plursisubharmonic function" 


The proof looks like first construct global PSH functions which are locally strictly PSH on small nbhd, then adding them together after multiplying some constants. I don't understand why that sum converges. Since the proof looks very straightforward and I may miss sth simple here. I would appreciate if anyone can help. Thanks! 



Holomorphic separation and the existence of strictly plurisubharmonic functions
I realized that my comment was misleading, so I prefer to put a complete answer. 


In Demailly's proof at some point you end up with a countable open cover $\{V\_j\}$ of your complex manifold $X$ and a family of non-negative smooth psh functions $\{v\_j\}$ such that $v\_j$ is strictly psh on $V\_j$. Then you want to sum them together in order to obtain a smooth non-negative strictly psh function $v$ on the whole of $X$.


At this point Demailly claims that you can choose a sequence of real numbers $\varepsilon\_j>0$ converging to zero so fast that 
$$
\sum\_{j=0}^{+\infty}\varepsilon\_jv\_j
$$
converges in the $C^{\infty}$-topology. This of course would suffice to get the desired function $v$, right?


I shall give you all the details for the $C^0$-convergence and I'll leave you the task to fill it out for the derivatives.


Fix an exhaustion by compact sets $\{K\_\nu\}$ of $X$ and set
$$
\alpha\_{j,\nu}:=\max\_{K\_\nu}|v\_j|.
$$
Now, for each $j\ge 1$ select a $\eta\_j>\max\_{1\le\nu\le j}\{\alpha\_{j,\nu}\}$. Finally, set $\varepsilon\_j:=1/(\eta\_j2^j)$.


To check uniform convergence on compacta note that for any given compact set $K\subset X$ you have that $K\subset K\_{\nu\_0}$ for some $\nu\_0$, so that, for $j\ge\nu\_0$, you have


$$
\max\_{K}|\varepsilon\_j v\_j|\le\max\_{K\_{\nu\_0}}|\varepsilon\_j v\_j|=\frac{\alpha\_{j,\nu\_0}}{\eta\_j 2^j}<\frac 1{2^j}.
$$


To obtain the $C^\infty$-convergence, you have to deal with the topology generated by all seminorms $p\_L^s$ when $s,L,\Omega$ vary, where $s\in\mathbb N$, $\Omega$ is a coordinate open set with coordinates $(x\_1,\dots,x\_m)$ and $L\subset \Omega$ is a compact subset. They are defined by
$$
p^s\_L(f)=\max\_{x\in L}\max\_{|\ell|\le s}|D^\ell f(x)|,
$$
where $\ell=(\ell\_1,\dots,\ell\_m)\in\mathbb N^m$, $|\ell|=\ell\_1+\cdots+\ell\_m$ and
$$
D^\ell f:=\frac{\partial^{|\ell|}f}{\partial x\_1^{\ell\_1}\cdots\partial x\_m^{\ell\_m}}.
$$ 
The way to define a sequence $\{\varepsilon\_j\}$ which works simultaneously for all derivatives is essentially the same, except for the flurry of subscripts... 


