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Question Number 131167 by 676597498 last updated on 02/Feb/21
Answered by SEKRET last updated on 02/Feb/21
Un=(1+1n2)(1+2n2)....(1+nn2)ln(Un)=ln(1+1n2)+ln(1+2n2)+..+(1+nn2)ln(Un)=∑nk=1ln(1+kn2)ln(Un)=ln((n2+1)nn2n)Un=(n2+1)nn2n
Commented by 676597498 last updated on 02/Feb/21
idontgetit
Answered by mathmax by abdo last updated on 02/Feb/21
un=∏k=1n(1+kn2)⇒vn=ln(un)=∑k=1nln(1+kn2)letprovefirstx−x22⩽ln(1+x)⩽xletf(x)=x−ln(1+x)withx⩾0f′(x)=1−11+x=x1+x⩾0⇒fisincreazinzbutf(0)=0⇒f(x)⩾0⇒x⩾ln(1+x)letg(x)=ln(1+x)−x+x22⇒g′(x)=11+x−1+x=1+x2−11+x=x21+x⩾0⇒gisincreazingbutg(o)=0⇒g(x)⩾0⇒x−x22⩽ln(1+x)⩽x⇒kn2−k22n4⩽ln(1+kn2)⩽kn2⇒1n2∑k=1nk−12n4∑k=1nk2⩽vn⩽1n2∑k=1nk⇒1n2×n(n+1)2−12n4n(n+1)(2n+1)6⩽vn⩽1n2×n(n+1)2⇒n2+n2n2−(n+1)(2n+1)12n3⩽vn⩽n2+n2n2⇒n+12n−(n+1)(2n+1)12n3⩽vn⩽n+12n⇒limn→+∞=12⇒limn→+∞ln(un)=12⇒limn→+∞un=e12=e
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