What are the partial limits of that series ?
tried for hours, couldn't figure this out. Thanks in advance.
Partial sums, you mean?
$\displaystyle \sum \Big(6(-1)^{n(n+1)/2} + 8 \cos ({n\pi\over 2})\Big) = 6\sum(-1)^{n(n+1)/2} + 8 \sum \cos ({n\pi\over 2})$. Suppose n is 0 or 3 mod 4. Then n(n+1)/2 is even. Otherwise, n(n+1)/2 is odd. This should tell you how to handle the first sum. Now suppose n is 0 mod 4. Then $\displaystyle \cos ({n\pi\over 2})$ is 1. If n is 2 mod 4, then it is -1. Otherwise, it is 0. This will let you handle the second sum.
Note that $\displaystyle \cos\left(\frac{n\pi}{2}\right)=\left\{\begin{arra y}{cl}0&,\,if\,\, n=1,3\!\!\!\pmod 4\\\!\!\!-1&,\,if\,\,n=2\!\!\!\pmod 4\\1&,\,if\,\,n=0\!\!\!\pmod 4\end{array}\right.$ , so :
$\displaystyle n=0\!\!\!\pmod 4\Longrightarrow a_n=14\,,\,\,n=1\!\!\!\pmod 4\Longrightarrow a_n=-6$ , $\displaystyle n=2\!\!\!\pmod 4\Longrightarrow a_n=-14\,,\,\,n=3\!\!\!\pmod 4\Longrightarrow a_n=6$ .
Check the above closely and proceed.
Tonio