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
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}}&amp;=\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}}&amp;=\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}
I would say that this limit goes to infinity. Under the condition that a least one of the $a_k &gt; 1$. The dividing by $n$ is is nothing compared to a power of $n$.