Infinite series

基本上這題的做法與http://hk.knowledge.yahoo.com/question/question?qid=7011113000518的類似。

希望今次不要再揀錯回答。

2011-12-16 08:32:17 補充：

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahooknowledge/infiniteseries/crazyinfiniteseries29.jpg

參考資料：
my wisdom of maths + result from http://www.wolframalpha.com/input/?i=x%28x%2B3%29%5E2-1%3D0 + formula from http://mathworld.wolfram.com/RiemannZetaFunction.html#eqn51

2011-12-18 08:17:36 補充：
自由自在，咁你玩唔玩得掂呢題？
http://tw.knowledge.yahoo.com/question/question?qid=1511121801486

2011-12-18 08:38:53 補充：
自由自在，你的做法雖好，但是似乎太不親民了。

相信冇乜人諗得明點解Σ(x = - ∞ to ∞)1/P(x) = - π Σ(k = 1 to n)Residue[cot πz/P(z),z_k]，其中P(x)是n次多項式。

My solution:
http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-2424.jpg

it cannot be solved.
a few properties can be found:
limit of the series exist.
the series is converging when k>0.

approximate solution: 0.854350537806538466580998019939560199550111875423327...