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Question Number 100785 by 175 last updated on 28/Jun/20
Answered by john santu last updated on 28/Jun/20
considerm⩽n3⩽m+1m3⩽n⩽m3+3m2+1∑∞n=1(−1)n+1⌊n⌋=∑∞m=1∑⌊n⌋=m(−1)n+1m=−∑∞m=1(−1)mm=−[∑∞m=1(−x)mm]x=0x=1=∫10∑∞k=0(−x)kdx=∫10dx1+x=[ln(x+1)]01=ln(2)
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