limit n tends to infinity for arbitrary non negative real numbers 
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
 A: 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}
A: 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 $$lim_{n\to\infty}N_n=a_m$$
A: 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:


*

*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.

*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.
A: 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 $n$.
