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Let $\{X_n\}_{n=1}^\infty$ be a convergence sequence such that $X_n \geq 0$ and $k \in \mathbb{N}$. Then $$ \lim_{n \to \infty} \sqrt[k]{X_n} = \sqrt[k]{\lim_{n \to \infty} X_n}. $$ Can someone help me figure out how to prove this?

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Hint: $$ a^p - b^p = (a-b)(a^{p-1}b^0 + \cdots + a^0b^{p-1} ) $$

details:

with $a = x_n^{1/p}, b = (\lim x_n)^{1/p}$ then $$ x_n - \lim x_n = ( x_n^{1/p}-(\lim x_n)^{1/p}) ( x_n^{(p-1)/p}(\lim x_n)^{0/p} + \cdots + x_n^{0/p} (\lim x_n)^{(p-1)/p} ) $$ The term $x_n^{(p-1)/p}(\lim x_n)^{0/p} + \cdots + x_n^{0/p} (\lim x_n)^{(p-1)/p} $ is $\ge$ $p\times \lim x_n/2$ when $n$ is big, because for such an $n$ $$ x_n \ge \lim x_n/2. $$

Hence $$ |x_n - \lim x_n| = |x_n^{1/p}-(\lim x_n)^{1/p}| ( x_n^{(p-1)/p}(\lim x_n)^{0/p} + \cdots + x_n^{0/p} (\lim x_n)^{(p-1)/p} ) \\ |x_n - \lim x_n| \ge p\times \lim x_n/2| x_n^{1/p}-(\lim x_n)^{1/p}| $$

When $n\to\infty$, LHS oges to zero, and so does $$ | x_n^{1/p}-(\lim x_n)^{1/p}| $$

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Assume uniqueness of solutions to $r = a^k$ when $k > 0$ and $a \geq 0$ and $r \geq 0$. Then you just need to prove that $(\lim_n a_n)^k = \lim_n a_n^k$. For this show that $f(x) = x^k$ is continuous, and the rest will follow.

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