Prove that lim(a₁ⁿ + a₂ⁿ)^1/n = a₁ as n goes to infinity?
Let a_1 ≥ a_2 > 0. Prove that lim(a₁ⁿ + a₂ⁿ)^1/n = a₁ as n goes to infinity. We can call this sequence {a_n}
The problem gives a hint; to apply the squeeze theorem. But to do so I need two more sequences, say {b_n} and {c_n} such that {b_n} ≤ {a_n} ≤ {c_n} don't I?
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Since a₁ ≥ a₂ > 0, we see that
(i) (a₁ⁿ + a₂ⁿ)^(1/n) ≤ (a₁ⁿ + a₁ⁿ)^(1/n) = 2^(1/n) a₁
(ii) (a₁ⁿ + a₂ⁿ)^(1/n) > (a₁ⁿ + 0)^(1/n) = a₁.
Hence, we can take b_n = a₁ and c_n = 2^(1/n) a₁ so that {b_n} ≤ {a_n} ≤ {c_n}.
Since lim {b_n} = a₁ = lim{c_n}, we conclude tby the Squeeze Theorem that
lim(a₁ⁿ + a₂ⁿ)^(1/n) = a₁.
I hope this helps!
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