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)] = …