All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 71649 by TawaTawa last updated on 18/Oct/19
Commented by mathmax by abdo last updated on 18/Oct/19
∑n=1[x]n=1+2+3+...+[x]23(x+12)3=23x3(1+12x)3=23x32(1+12x)3∼23x32letfindan∼forf(x)=∑k=1[x]kwehave1+2+3+...+[x]=[x]{1[x]+2[x]+3[x]+...+[x]−1[x]}[x]=N=[N]{∑k=1N−1kN}=NN×1N∑k=1N−1kNbut1N∑k=1N−1kN=∫01xdx=23x33]01=23⇒1+2+3+...+[x]∼23[x][x]=23[x]32∼23x32limx→+∞(23(x+12)3−∑n=1xn)=0andifyouhavetheanswerpostit.
Commented by TawaTawa last updated on 19/Oct/19
Idon′thavetheanswersir.Iwilluseyouranswer.Godblessyousir.
Commented by Abdo msup. last updated on 19/Oct/19
youaremostwelcome.
Terms of Service
Privacy Policy
Contact: info@tinkutara.com