Infinite series(difficult)

N Σ (2n - 1)^3 n=1 is the sum of the cubes of the first N odds. There are a couple different approaches to evaluating this. I think perhaps the simplest is to take the sum of the first 2N - 1 cubes, and subtract out the first N-1 even cubes. Note that 2N-1 Σ n^3 = (2N-1)²(2N)²/4 = N²(2N-1)² = N²(4N² - 4N + 1). n=1 The sum of the first N - 1 even cubes is N-1 Σ (2n)^3 = 8(N-1)²(N)²/4 = 2N²(N² - 2N + 1) n=1 So subtracting gives the sum N Σ (2n-1)^3 = N²(4N² - 4N + 1) - 2N²(N² - 2N + 1) = 2N^4 - N². n=1

It's so sorry! I chose a wrong solution. The correct answer is doraemonpaul's.

圖片參考：http://i212.photobucket.com/albums/cc82/doraemonpaul/yahooknowledge/infiniteseries/goodinfiniteseries9.jpg

參考資料：
my wisdom of maths + formula from http://mathworld.wolfram.com/RiemannZetaFunction.html#eqn51

2011-12-13 11:18:42 補充：
這題最困難的地方就是若直接把1/(8k³ + 1)進行部分分式會分拆出發散級數，因為Σ(k = a to ∞)P(k)/Q(k)，其中deg(Q(k)) - deg(P(k)) ≤ 1，是恆為發散的。

因此我嘗試把兩邊的無限級數拆開，結果我砌出只有偶次項的有理函數級數，把它進行部分分式就可以分拆出收歛且較低次項的有理函數級數，問題就自然迎刃而解了。

2011-12-15 17:17:06 補充：
CRebecca，你是否有病？竟然選擇rio******那個垃圾回答做最佳解答？

π/6*[1+√3 sinh(√3 π/2)]/cosh(√3 π/2)