摘 要: | 本文利用初等方法证明∑∞n =11n4 =π490 .1 几个引理引理 1 ∑∞n =1cot2 nπ2m+1 =13 m(2m-1 ) ,∑mn ,l =1n<lcot2 nπ2m +1 cot2 lπ2m +1=13 0 m (m -1 ) (2m -2 ) (2m-3 ) .其中m、l、n等均表示整数 ,下同 .证明 由de·Movre公式得cos(2m +1 )α+isin(2m +1 )α=(cosα+isinα) 2m+1于是 ,cos(2m +1 )α+isin(2m+1 )α=∑mk =0(-1 ) kC2k2m+1cos2 (m-k) +1αsin2kα+i∑mk =0(-1 ) kC2k+12m+1cos2 (m-k) αsin2k+1α. (1 )比…
|