limit n tends to infinity for arbitrary non negative real numbers

Question

I tried evaluating the limit taking log and afterwards i couldnt proceed further. I tried solving using numbers instead of variables and did not got any relation from that and help in this matter is appreciated

Answer 1

Let $A=\max\{a_1,a_2,\ldots,a_k\}$, then \begin{align} \lim_{n\to\infty}(a_1^n+a_2^n+\ldots+a_k^n)^{\frac{1}{n}}&=\lim_{n\to\infty} A\left[\left(\frac{a_1}{A}\right)^n+\left(\frac{a_2}{A}\right)^n+\ldots+\left(\frac{a_k}{A}\right)^n\right]^{\frac{1}{n}}\\ \end{align} Since $$A\leq A\left[\left(\frac{a_1}{A}\right)^n+\left(\frac{a_2}{A}\right)^n+\ldots+\left(\frac{a_k}{A}\right)^n\right]^{\frac{1}{n}} \le Ak^{\frac{1}{n}}$$ And $$\lim_{n\to \infty} k^{1/n}=\exp\left(\lim_{n\to\infty}\frac{\ln k}{n}\right)=\exp(0)=1$$ It follows, from the sandwich theorem, that \begin{align} \lim_{n\to\infty}(a_1^n+a_2^n+\ldots+a_k^n)^{\frac{1}{n}}&=\lim_{n\to\infty} A\left[\left(\frac{a_1}{A}\right)^n+\left(\frac{a_2}{A}\right)^n+\ldots+\left(\frac{a_k}{A}\right)^n\right]^{\frac{1}{n}}=A\\ \end{align}

Answer 2

Let $a_m$ be Max {$a_1,a_2,.....,a_k$}. We have

$N_n=(a_m^n)^{1/n}(M+r_1+r_2+...+r_j)^{1/n}=a_m(M+r_1+r_2+...+r_j)^{1/n}$ where $M$ is the number of times $a_m$ is repeated among the $a_i's$ (obviously $M$ can be equal to $1$) and the fractions $r_j$ are less than $1$.

Hence $N_n\to a_m(M+R)^{1/n}$ where $R\to 0$.

Thus $$lim_{n\to\infty}N_n=a_m$$

Answer 3

I would say that this limit goes to infinity. Under the condition that a least one of the $a_k > 1$. The dividing by $n$ is is nothing compared to a power of $n$.

Answer 4

Here is a version with L'Hôpital's rule. It's not quite as strong (logically) as the other answers, as we'll explain.

Assume that $a_1 \leq a_2 \leq \dots \leq a_k$. Starting from $$ L = \lim_{n\to\infty} \left(a_1^n + a_2^n + \dots + a_k^n\right)^{1/n} $$ we know $$\begin{split} \ln L &= \ln \lim_{n\to\infty} \left(a_1^n + a_2^n + \dots + a_k^n\right)^{1/n} \\ &= \lim_{n\to\infty}\ln\left(a_1^n + a_2^n + \dots + a_k^n\right)^{1/n} \\ &= \lim_{n\to\infty}\frac{\ln\left(a_1^n + a_2^n + \dots + a_k^n\right)}{n}\\ &\stackrel{\text{H}}{=} \lim_{n\to\infty}\frac{(\ln a_1) a_1^n + (\ln a_2)a_2^n + \dots + (\ln a_k)a_k^n}{a_1^n + a_2^n + \dots + a_k^n} \\ &= \lim_{n\to\infty} \frac{a_k^n}{a_k^n}\cdot \frac{(\ln a_1) \left(\frac{a_1}{a_k}\right)^n + (\ln a_2)\left(\frac{a_2}{a_k}\right)^n + \dots + (\ln a_{k-1})\left(\frac{a_{k-1}}{a_k}\right)^n+ \ln a_k}{\left(\frac{a_1}{a_k}\right)^n + \left(\frac{a_2}{a_k}\right)^n + \dots + \left(\frac{a_{k-1}}{a_k}\right)^n+1} \\ &= 1 \cdot \frac{0 + 0 + \dots + 0 + \ln a_k}{0 + 0 + \dots + 0 + 1} = \ln a_k \end{split}$$ (“H” denotes the invocation of L'Hôpital's rule). Therefore $L=a_k$.

Remarks:

1. Limit proofs with L'Hôpital's rule are “backwards” in the sense that we don't know that the limit $L$ exists until we get to the end.

2. To verify the power sequence is indeterminate requires a check of cases. If $a_k > 1$ then the base $a_1^n + \dots + a_k^n$ tends to $\infty$ so the limit is of the indeterminate form $\infty^0$. If $a_k < 1$ then the base $a_1^n + \dots + a_k^n$ tends to $0$ so the limit is of the indeterminate form $0^0$. If $a_k =1$, the base $a_1^n + \dots + a_k^n$ tends to $1$. This form ($1^0$) is not indeterminate; the limit is $1$, but that requires a separate proof without L'Hôpital's rule.