Question 4.T.9: For any positive sequence (xn), lim sup n^√xn ≤ lim sup lim ...
For any positive sequence (x_{n}),
\lim \sup \sqrt[n]{x_{n}} ≤ \lim \sup \frac{x_{n+1}}{x_{n}}
\lim \inf \frac{x_{n} +1}{x_{n}} ≤ \lim \inf \sqrt[n]{x_{n}}.
To prove the first inequality, let
L = \lim \sup \frac{x_{n} +1}{x_{n}}.
If L = ∞ there is nothing to prove. If 0 ≤ L < ∞ then, for any M > L, there is a positive integer N such that
\frac{x_{n} +1}{x_{n}} ≤ M for all n ≥ N.
In particular, for any k ∈ \mathbb{N},
x_{N+k} ≤ Mx_{N+k−1}
≤ M^{2}x_{N+k−2}
≤ · · ·
≤ M^{k}x_{N}.
Hence
n > N ⇒ x_{n} ≤ M^{n−N}x_{N}
⇒ \sqrt[n]{x_{n}} ≤ M \sqrt[n]{x_{N}M^{−N}}.
From Example 3.8, \lim \sqrt[n]{x_{N}M^{−N}} = 1, and from Exercise 3.6.1 and Theorem 3.16(iii) we obtain
\lim \sup \sqrt[n]{x_{n}} ≤ M.
Since this is true for any M > L, it follows that \lim \sup \sqrt[n]{x_{n}}≤ L.
The second inequality is proved by a similar procedure.