下面这几题的敛散性,和级数的和怎么求?∑(1⼀√n-1⼀√n+1), ∑(-1)^n-1⼀3^n, ∑1⼀a^2n-1

2025-05-22 20:52:59
推荐回答(1个)
回答1:

  1)由
  Sn = ∑(k=1~n)[1/√k - 1/√(k+1)]
   = 1 - 1/√(n+1)
  → 1 (n→inf.),
得知∑(n=1~inf.)[1/√n - 1/√(n+1)] = 1。

  2)由
  Sn = ∑(k=1~n)[(-1)^(k-1)]/3^k
   = (1/3)∑(k=1~n)[(-1/3)^(k-1)]
   = (1/3)[1-(-1/3)^n]/[1-(-1/3)]
   → (1/3)(3/4)= 1/4 (n→inf.),
得知
    ∑(n=1~inf.)[(-1)^(n-1)]/3^n = 1/4。

  3)由
  Sn = ∑(k=1~n)[1/a^(2k-1)]
   = (1/a)*∑(k=1~n)[(1/a^2)]^(k-1)
   = … (同2))
得知
  ∑(n=1~inf.)[1/a^(2n-1)] = …