3
$\begingroup$

Let $a_1, \ldots, a_n > 0$. How to compute the limit $\lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n}$?

My solution: $$ \lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n} \\ = \lim_{n \to \infty} (1 + a_1^n + \cdots + a_m^n - 1)^{1/n} \\ = \lim_{n \to \infty} \left( (1 + a_1^n + \cdots + a_m^n - 1)^{\frac{1}{a_1^n + \cdots + a_m^n - 1}} \right)^{ \frac{a_1^n + \cdots a_m^n-1}{n}}. $$ But then we need to have the condition $a_1^n + \cdots + a_m^n - 1 \to 0$ ($n \to \infty$).

Thank you very much.

$\endgroup$
3
$\begingroup$

Let $A = \max \{a_1, \cdots, a_m\}$. Then $$\left( a_1^n + \cdots + a_m^n \right)^{\frac 1n} \ge (A^n)^\frac{1}{n} = A.$$ On the other hand, $$\left( a_1^n + \cdots + a_m^n \right)^{\frac 1n}\le (A^n + A^n + \cdots + A^n)^{\frac 1n} = m^{\frac 1n} A$$ This implies

$$\liminf \left( a_1^n + \cdots + a_m^n \right)^{\frac 1n} \le A.$$

So the limit is $A$.

Remark: Note that $$\left( a_1^n + \cdots + a_m^n \right)^{\frac 1n}$$

is the $n$-th norm of the vector $v =(a_1, a_2, \cdots, a_m)$. The above shows that

$$ \lim_{n \to \infty} \|v\|_n = \| v\|_\infty = \max\{ a_1, \cdots, a_m\}.$$

This has a generalization in measure theory, concerning the limit of the $L^p$ norm of a function $f$.

$\endgroup$
  • 1
    $\begingroup$ It reminds me of this exercise; If $f:[0,1]\to R$ is continuous then $\lim_{n \to\infty}(\int_0^1 |f(x)|^n dx)^{1/n}=\max \{|f(x)| : x\in [0,1]\}$. $\endgroup$ – DanielWainfleet Oct 26 '15 at 5:42
  • $\begingroup$ @user254665 : Exactly. See my remark there. $\endgroup$ – user99914 Oct 26 '15 at 5:43
  • 1
    $\begingroup$ @JohnMa Just curious, but it also seems that $\displaystyle\lim_{x\to -\infty} \sqrt[n]{{a_1}^n+\cdots +{a_m}^n}=\min\{a_1,\cdots ,a_m\}$. Is this correct? (if so, I'm sure it can be proven analogously) $\endgroup$ – Corellian Oct 27 '15 at 1:55
  • 1
    $\begingroup$ @JohnMa Of course, I meant to write *$\displaystyle\lim_{n\to -\infty}$. $\endgroup$ – Corellian Oct 27 '15 at 3:18
  • 1
    $\begingroup$ I think that's true: If $n<0$, we have $(a_1^n + \cdots a_m^n)^\frac{1}{n} = \frac{1}{((\frac{1}{a_1})^{|n|} + \cdots + (\frac{1}{a_1})^{|n|} )^\frac{1}{|n|}} \to \frac{1}{ \max \{ \frac{1}{a_1}, \cdots , \frac{1}{a_m}\}} = \min\{a_1, \cdots, a_m\}$ @Brody $\endgroup$ – user99914 Oct 27 '15 at 3:32
2
$\begingroup$

Think about the part under the root. The largest term will become more and more significant compared to the others. In the limit it will completely dominate the other terms. The root effectively undoes the power so the function is equivalent to the maximum function.

$$\lim_{n \to \infty} \sqrt[n]{a_1^n + \cdots + a_m^n}=\max(a_1,a_2,\cdots,a_m)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.