find a $B_{n,j}$ such that $|A_{n,j}-L_j| \leq B_{n,j}$ $\forall n,j$ and $\sum_{j=0}^{\infty}B_{n,j}$ converges We have $A_{n,j}= 3(-1)^j2^{n-j+1}\frac{(2(n-j)-4)!}{(n-j)!(n-j-2)!}\binom{j+2}{2}\frac{n^\frac{5}{2}}{8^n}$
and $L_j=(-\frac{1}{8})^j\binom{j+2}{2}\frac{3}{8\sqrt{\pi}}$
So I know $\lim_{n \to \infty} A_{n,j} = L_j$ implies $\lim_{n \to \infty}  \sum_{j=0}^{1000} A_{n,j}=\sum_{j=0}^{1000}L_j $.
Now I want to prove that $\lim_{n \to \infty}  \sum_{j=0}^{n} A_{n,j}=\sum_{j=0}^{\infty}L_j $
So I was thinking that I want to show that $\forall$ small $\epsilon > 0, \exists$ large $N$ such that $   |\sum_{j=0}^{N} A_{N,j}-\sum_{j=0}^{\infty}L_j | < \epsilon$.
I know that $|\sum_{i=0}^{N} A_{N,j}-\sum_{j=0}^{\infty}L_j | \leq \sum_{j=0}^{N}|A_{N,j}-L_j|+|\sum_{j=N+1}^{\infty}L_j|$. Since $|\sum_{j=N+1}^{\infty}L_j|$ converges say to less than $\frac{\epsilon}{2}$, we need to show that $\sum_{j=0}^{N}|A_{N,j}-L_j|$ is less than $\frac{\epsilon}{2}$.
This is similar to the Wierstrauss M-test, I would like to find a $B_{j}$ such that $|A_{n,j}-L_j| \leq |A_{n,j}| \leq B_{j}$  $\forall n,j$ and  $\sum_{j=0}^{\infty}B_{j}$ converges.
How do I find such $B_{j}$? Need some help.  
 A: Let us start by putting things in a nicer form.
$$\begin{eqnarray*}A_{n,j}&=&3(-1)^j 2^{n-j+1}\frac{(2n-2j-4)!}{(n-j)!(n-j-2)!}\binom{j+2}{2}\frac{n^{5/2}}{8^n}\\&=&\frac{6\cdot 2^{n-j}}{8^n}(-1)^j\frac{(n-j)(n-j-1)n^{5/2}}{(2n-2j)(2n-2j-1)(2n-2j-2)(2n-2j-3)}\binom{2n-2j}{n-j}\binom{j+2}{2}\end{eqnarray*}$$
but due to Stirling's formula,
$$\binom{2m}{m}=\frac{4^{m}}{\sqrt{\pi m}}\left(1+O\left(\frac{1}{m}\right)\right),$$
hence:
$$\begin{eqnarray*}A_{n,j}&=&\frac{6\cdot 2^{n-j}}{16\cdot 8^n}(-1)^j \frac{n^{5/2}}{(n-j)^2}\,2^{2n-2j}\frac{1}{\sqrt{\pi(n-j)}}\binom{j+2}{j}\left(1+O\left(\frac{j}{n}\right)\right)\\&=&\frac{3(-1)^j}{8^{j+1}\sqrt{\pi}}\left(\frac{n}{n-j}\right)^{5/2}\binom{j+2}{2}\left(1+O\left(\frac{j}{n}\right)\right)\\&=& L_j \left(1+O\left(\frac{j}{n}\right)\right)\end{eqnarray*}$$
and:
$$ |A_{n,j}-L_j|\leq K|L_j|\frac{j}{n} $$
for an absolute constant $K>0$. Since obviously $\sum |L_j|=O(1)$,
$$\left|\sum_{j=0}^{N} A_{N^2,j}-\sum_{j=0}^{N} L_j\right|\ll\frac{K}{N}$$
but you have to be smart enough to choose the "approximation parameter" $n$ in such a way that it is substantially bigger than the length of the partial sum you are considering. As shown, with the choice $n=N^2$ everything works just fine.
