设0<x<1,证明∑从0到∞ xⁿ（1-x）ⁿ≤256/229
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解决时间 2021-03-24 21:07
- 提问者网友：听门外雪花风
- 2021-03-24 15:32
设0<x<1,证明∑从0到∞ xⁿ（1-x）ⁿ≤256/229
最佳答案
- 五星知识达人网友：风格不统一
- 2021-03-24 16:52
(1)求1+x⁴+x⁸+……+x⁴ⁿ的和，-1
当n→+∞limS‹n›=S=n→+∞lim(1-x⁴ⁿ)/(1-x⁴)=1/(1-x⁴) (∵-1
S′‹n›=x⁴+x⁸+x¹²+……+x⁴ⁿ=x⁴(1-x⁴ⁿ)/(1-x⁴)
n→+∞limS′‹n›=S′=x⁴/(1-x⁴)
故S=∫[x⁴/(1-x⁴)]dx=∫[-1+1/(1-x⁴)]dx=-x+∫dx/(1-x⁴)=-x+∫dx/(1+x²)(1-x²)
=-x+(1/2)∫[1/(1+x²)+1/(1-x²)]dx=-x+(1/2)[∫dx/(1+x²)+∫dx/(1-x²)]=-x+(1/2)[arctanx-∫dx/(x²-1)]
=-x+(1/2)arctanx-(1/4)ln[(x-1)/(x+1)]+C
当n→+∞limS‹n›=S=n→+∞lim(1-x⁴ⁿ)/(1-x⁴)=1/(1-x⁴) (∵-1
S′‹n›=x⁴+x⁸+x¹²+……+x⁴ⁿ=x⁴(1-x⁴ⁿ)/(1-x⁴)
n→+∞limS′‹n›=S′=x⁴/(1-x⁴)
故S=∫[x⁴/(1-x⁴)]dx=∫[-1+1/(1-x⁴)]dx=-x+∫dx/(1-x⁴)=-x+∫dx/(1+x²)(1-x²)
=-x+(1/2)∫[1/(1+x²)+1/(1-x²)]dx=-x+(1/2)[∫dx/(1+x²)+∫dx/(1-x²)]=-x+(1/2)[arctanx-∫dx/(x²-1)]
=-x+(1/2)arctanx-(1/4)ln[(x-1)/(x+1)]+C
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