# ${a_n}^{1/k} \rightarrow m^{1/k}$

Let $k$ be a fixed positive natural number and $\{a_n\}$ be a sequence of positive reals converging to $m$ then show that

$$(a_n)^{1/k} \hspace{.1cm} \text{converges to } \hspace{.1cm} m^{1/k}$$

as $n$ goes to $\infty$.

My attempt: I need a hint to start it.

• The function $x \mapsto x^{1/k}$ is continuous on $[0,\infty)$. Do you know a general theorem about continuity and sequences? – Bungo Aug 12 '15 at 16:55
• no didnt get it – ketan Aug 12 '15 at 17:01
• I guess probably the OP needs a direct proof. – Zhanxiong Aug 12 '15 at 17:04

Take a look at Lemma 1, here. The gist is that, for any sequence $a_n$, if $f(x)$ is continuous on your domain, then $$\lim_{n\to \infty}f(a_n) = f(\lim_{n\to \infty}\ a_n)$$

If one wishes to use the direct definition of a limit with $\epsilon$, then the proof would be a bit technical.

Note that this argument below works only if $k$ is a positive natural number, which is what you have. If $k$ is given to be only a non-zero real number, then it is best to use the definition of a continuous map as outlined in the answer by @rwbogl.

Proof: We have $a_n \to m$ as $n \to \infty$. For all $\epsilon > 0$, there exists $N$ such that $|a_n-m|<\epsilon$ if $n \ge N$. Now, we have \begin{align*} |a_n^{\frac 1k}-m^{\frac 1k}|&=\frac{|a_n-m|}{\left|a_n^{\frac{k-1}k}+a_n^{\frac{k-2}k}m^{\frac 1k}+a_n^{\frac{k-3}k}m^{\frac 2k}+\cdots+a_n^{\frac 2k}m^{\frac{k-3}k}+a_n^{\frac 1k} m^{\frac{k-2}k}+m^{\frac{k-1}k} \right|} \\ &< \frac{\epsilon}{\left|a_n^{\frac{k-1}k}+a_n^{\frac{k-2}k}m^{\frac 1k}+a_n^{\frac{k-3}k}m^{\frac 2k}+\cdots+a_n^{\frac 2k}m^{\frac{k-3}k}+a_n^{\frac 1k} m^{\frac{k-2}k}+m^{\frac{k-1}k} \right|}. \end{align*} Thus, $a_n^{\frac 1k} \to m^{\frac 1k}$ as $n \to \infty$. $\blacksquare$

This argument works because the denominator $$\left|a_n^{\frac{k-1}k}+a_n^{\frac{k-2}k}m^{\frac 1k}+a_n^{\frac{k-3}k}m^{\frac 2k}+\cdots+a_n^{\frac 2k}m^{\frac{k-3}k}+a_n^{\frac 1k} m^{\frac{k-2}k}+m^{\frac{k-1}k} \right|$$ is non-zero and bounded. The denominator is bounded because each $a_n$ is bounded (because the sequence $\{a_n\}$ is convergent by hypothesis), $m$ is a finite number, and the denominator contains only a finite number of terms. Also, the denominator is non-zero since each $a_n$ is positive by hypothesis; this means precisely that at least the first term, $a_n^{\frac{k-1}k}$, is non-zero (all the other terms in the denominator would be zero if the limit of $\{a_n\}_{n=1}^\infty$ is zero, that is, $m=0$).

Here is a proof without using continuity.

If $m = 0$, then it needs to be shown that $a_n^{1/k} \to 0$, which follows from $$|a_n^{1/k} - 0| = |a_n|^{1/k} = |a_n - 0|^{1/k}.$$

If $m > 0$, let $b_n = a_n^{1/k}$ and $M = m^{1/k}$, then $M > 0$, it then follows that $$|a_n - m| = |b_n^k - M^k| = |b_n - M|(b_n^{k - 1} + b_n^{k - 2}M + \cdots + M^{k - 1}) \geq |b_n - M|M^{k - 1}$$ Hence $$|b_n - M| \leq \frac{|a_n - m|}{M^{k - 1}}$$ and the result easily follows.

• $|b_n - M|M^{k - 1}$ should be $k|b_n - M|\min(M, b_n)^{k - 1}$. – marty cohen Aug 20 '15 at 5:49
• $|b_n - M|(b_n^{k - 1} + \cdots + M^{k - 1}) - |b_n - M|M^{k - 1} = |b_n - M|(b_n^{k - 1} + \cdots + b_n M^{k - 2}) \geq 0$, I didn't see any problem of my inequality above. – Zhanxiong Aug 21 '15 at 4:01
• I see. You are throwing away k-1 of the terms to get a lower bound. That's OK. – marty cohen Aug 21 '15 at 4:49

If $f$ is continuous then $x_n \to x \Rightarrow f(x_n) \to f(x)$

To prove that $f(x) = x^{1/k}$ is continuous, note that its inverse $g(y) = y^k$ is continuous so since $x_n \to m$ implies $y_n = f(x_n)$ is bounded one can consider a converging subsequence $y_{n_k} \to y$ now, since

$$g(y_{n_k}) = x_{n_k} \to m \Rightarrow g(y) = m \Rightarrow y = m^{1/k}$$

This proves that the only limit poiny of the bounded sequence $y_n$ is $m^{1/k}$. Therefore the sequence $y_n$ converges to $m^{1/k}$

• any alternative? – ketan Aug 12 '15 at 17:07
• ketan, what do you mean by alternative? – Conrado Costa Aug 12 '15 at 17:08
• alternative way to do. continuity concept used here is alien to me – ketan Aug 12 '15 at 17:09
• continuity is a crucial concept in your question, it means exactly that on the real case :that converging sequences are mapped into converging sequences, however you can consider $|y_n - m^{1/k}|> \delta$ and note that $y_n^k - m> \epsilon$ – Conrado Costa Aug 12 '15 at 17:14