I have to prove that $u_n=4^{3n}/3^{4n}$ is a null sequence. My attempt: $1/3^{4n}<4^{3n}/3^{4n}<1/3^{n/4}$ Taking limits and by squeeze principle the limit $u_n=4^{3n}/3^{4n}$ tends to 0. Is this correct?
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I don't know how you got your last inequality. Why don't you just write your sequence as $$u_n=\frac{4^{3n}}{3^{4n}}=\left(\frac{4^3}{3^4}\right)^n=\left(\frac{64}{81}\right)^n$$ Since $0<64/81<1$, the sequence tends to $0$.
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1$\begingroup$ math.stackexchange.com/questions/2400996/is-nᵐmⁿ-if-mn/… $\endgroup$ Commented Apr 16, 2021 at 14:29
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$\begingroup$ I thought about that but is it accepted as a proof? $\endgroup$ Commented Apr 16, 2021 at 14:31
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$\begingroup$ You can do it that way, in the general case, but calculating $3^4$ and $4^3$ is trivial $\endgroup$– AndreiCommented Apr 16, 2021 at 14:37
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