Understanding Borel transform (sum) in QM

[Karlisbad]

Hello, i have heard tat you can use Borel transform in QM so the divergent perturbation series become convergent.. i am a bit confused since if you must compute the Borel transform:

$$B(x)=\sum_{n=0}^{\infty}\frac{a(n)x^{n}}{n!}$$

But this can only be made for a few cases $$a(n)=(-1)^{n}$$ and
$$a(n)=(-1)^{n} n!$$

But if the a(n) don't follow a known pattern Borel Transform is useless..

 

[Karlisbad]

sorry the first "image" (i really hate this f... Latex ) should be:

$$\sum_{n=0}^{\infty}(a(n) x^{n})/n!$$

-the second: $$a(n)= (-1)^{n} n!$$

- And the third $$a(n)= (-1)^{n}$$