无穷级数和的计算

lin2009

请先运行下面的语句，
sum((1/9)*sin((1/6*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2/(sin((1/18*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2*(3*k-3/2+(1/2)*(-1)^k)^2), k = 1 .. 10)
sum((1/9)*sin((1/6*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2/(sin((1/18*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2*(3*k-3/2+(1/2)*(-1)^k)^2), k = 1 .. 20)
sum((1/9)*sin((1/6*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2/(sin((1/18*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2*(3*k-3/2+(1/2)*(-1)^k)^2), k = 1 .. 30)
sum((1/9)*sin((1/6*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2/(sin((1/18*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2*(3*k-3/2+(1/2)*(-1)^k)^2), k = 1 .. 40)
sum((1/9)*sin((1/6*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2/(sin((1/18*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2*(3*k-3/2+(1/2)*(-1)^k)^2), k = 1 .. 50)

然后求出 k = 1 .. infinity各项之和的值（如下，所用的函数或方法不限），给出过程和结果。
sum((1/9)*sin((1/6*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2/(sin((1/18*(3*k-3/2+(1/2)*(-1)^k))*Pi)^2*(3*k-3/2+(1/2)*(-1)^k)^2), k = 1 .. infinity)

junl06

数值解0.9341600873459063713300135514286068975811320426606509646399466487780289338092251876092213374306936942
同样可以得到解析解，比较复杂，可以看最下面，求解过程如下：
首先观察V=3*k-3/2+(1/2)*(-1)^k），可知这是如下数列：1,5,7,11,13,17,19,23,25,29,31,35,37.。。。（相邻两数的差分别为4,2，递推）
考虑到sin函数的周期是2*pi，算式中出现的两个sin函数中pi的分母有最小公倍数18，同时36是上边数列的周期。
最终确定求解方式，每12个数作为一组，每组都是(1/9)*S，S内容如下
-(1/(36*k+1)^2+1/(36*k+17)^2+1/(36*k+19)^2+1/(36*k+35)^2)/(2*(cos((1/9)*Pi)-1))
+(1/(36*k+5)^2+1/(36*k+13)^2+1/(36*k+23)^2+1/(36*k+31)^2)/(2*(1+cos((4/9)*Pi)))
+(1/(36*k+7)^2+1/(36*k+11)^2+1/(36*k+25)^2+1/(36*k+29)^2)/(2*(1+cos((2/9)*Pi)))
这里面-1/(2*(cos((1/9)*Pi)-1))是V=1,17,19,35的时候sin(pi*V/6)^2/sin(pi*V/18)^2化简结果，由于相同进行了合并同类项，同理对V=5,13,23,31合并，V=7,11,25,29合并。
用sum函数，k=0~无穷 求和，就可以得到结果。代码如下，要看解析解，可以把evalf函数去掉再运行

evalf((1/9)*(sum(-(1/(36*k+1)^2+1/(36*k+17)^2+1/(36*k+19)^2+1/(36*k+35)^2)/(2*(cos((1/9)*Pi)-1))+(1/(36*k+5)^2+1/(36*k+13)^2+1/(36*k+23)^2+1/(36*k+31)^2)/(2*(1+cos((4/9)*Pi)))+(1/(36*k+7)^2+1/(36*k+11)^2+1/(36*k+25)^2+1/(36*k+29)^2)/(2*(1+cos((2/9)*Pi))), k = 0 .. infinity)))

lin2009

数值解0.9341600873459063713300135514286068975811320426606509646399466487780289338092251876092213374306936942
同样可以得到解析解，比较复杂，可以看最下面，求解过程如下：
首先观察V=3*k-3/2+(1/2)*(-1)^k ...
junl06 发表于 2011-3-9 15:47

怎么会有多种结果呢？（maple14.01）
> evalf((1/9)*(sum(-(1/(36*k+1)^2+1/(36*k+17)^2+1/(36*k+19)^2+1/(36*k+35)^2)/(2*(cos((1/9)*Pi)-1))+(1/(36*k+5)^2+1/(36*k+13)^2+1/(36*k+23)^2+1/(36*k+31)^2)/(2*(1+cos((4/9)*Pi)))+(1/(36*k+7)^2+1/(36*k+11)^2+1/(36*k+25)^2+1/(36*k+29)^2)/(2*(1+cos((2/9)*Pi))), k = 0 .. infinity)));
print(`output redirected...`); # input placeholder
                        -Float(infinity)

> evalf((1/9)*(sum(-(1/(36*k+1)^2+1/(36*k+17)^2+1/(36*k+19)^2+1/(36*k+35)^2)/(2*(cos((1/9)*Pi)-1))+(1/(36*k+5)^2+1/(36*k+13)^2+1/(36*k+23)^2+1/(36*k+31)^2)/(2*(1+cos((4/9)*Pi)))+(1/(36*k+7)^2+1/(36*k+11)^2+1/(36*k+25)^2+1/(36*k+29)^2)/(2*(1+cos((2/9)*Pi))), k = 0 .. infinity)));
print(`output redirected...`); # input placeholder
                        0.00006769275995
S = > (1/9)*(sum(-(1/(36*k+1)^2+1/(36*k+17)^2+1/(36*k+19)^2+1/(36*k+35)^2)/(2*(cos((1/9)*Pi)-1))+(1/(36*k+5)^2+1/(36*k+13)^2+1/(36*k+23)^2+1/(36*k+31)^2)/(2*(1+cos((4/9)*Pi)))+(1/(36*k+7)^2+1/(36*k+11)^2+1/(36*k+25)^2+1/(36*k+29)^2)/(2*(1+cos((2/9)*Pi))), k = 0 .. infinity))

(1/729)*Pi^2*(-738385703904*sqrt(3)*cos((1/18)*Pi)^5+157527782880*sqrt(3)*cos((1/18)*Pi)^3+862509929280*sqrt(3)*cos((1/18)*Pi)^4+592200138240*sqrt(3)*cos((1/18)*Pi)^8-1184400276480*sqrt(3)*cos((1/18)*Pi)^6+583203312000*sqrt(3)*cos((1/18)*Pi)^7+158322384930*sqrt(3)*cos((1/18)*Pi)+440752937616*cos((1/18)*Pi)-253730359140*cos((1/18)*Pi)^2-3461453740752*cos((1/18)*Pi)^3+2962337598720*cos((1/18)*Pi)^4+7643019320832*cos((1/18)*Pi)^5-6153691623424*cos((1/18)*Pi)^6-4641767931648*cos((1/18)*Pi)^7+3076845811712*cos((1/18)*Pi)^8-343297034160*cos((1/18)*Pi)^2*sqrt(3)+56402654700*sqrt(3)+97692185661)/(-32800663177368*cos((1/18)*Pi)-4465545430532*sqrt(3)-15327403794792*cos((1/18)*Pi)^2+321446404953088*cos((1/18)*Pi)^8+307880365985424*cos((1/18)*Pi)^4-91501100991488*cos((1/18)*Pi)^7-642892809906176*cos((1/18)*Pi)^6+27228550139200*cos((1/18)*Pi)^5+86059507377328*cos((1/18)*Pi)^3+5217066394181+3575834196306*sqrt(3)*cos((1/18)*Pi)+102864998558720*sqrt(3)*cos((1/18)*Pi)^8-205729997117440*sqrt(3)*cos((1/18)*Pi)^6+123893282023872*sqrt(3)*cos((1/18)*Pi)^4-11661918545240*cos((1/18)*Pi)^2*sqrt(3)-205724556095072*sqrt(3)*cos((1/18)*Pi)^5+87958599237880*sqrt(3)*cos((1/18)*Pi)^3+128912073052160*sqrt(3)*cos((1/18)*Pi)^7)

> evalf(S)
-0.6092348396e-5

junl06

3# lin2009

确实有这种情况，把显示精度改成100位。
具体原因我也不太清楚，用的是MAPLE11。

把原题中的k，分别计算到480,600,720,800，等，可以看到是逼近于上面的数值解，应该没错。print(`output redirected...`)，这句我没用使用。