Indian Statistical Institute B.Math & B.Stat : Limits at Infinity

Let $a_1 > a_2 > \dots > a_r$ be positive real numbers. Compute $\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}}$. Since $a_1 > a_2 > \dots > a_r$ and each of them is positive we have $a_1^n>a_2^n>\dots>a_r^n$ $\implies a_1^n+a_2^n+\dots+a_r^n < a_1^n+a_1^n+\dots+a_1^n = ra_1^n$ Letting $n \to \infty$ we have, $\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} < \lim_{n \to \infty}(ra_1^n)^{\frac{1}{n}}$ $= a_1\lim_{n \to \infty}r^{\frac{1}{n}}= a_1$ Note $r>0$ Now, $\big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} = \bigg( a_1^n \big(1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1}\big) \bigg)^{\frac{1}{n}} = a_1\bigg( 1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1}\big) \bigg)^{\frac{1}{n}} > a_1$ Since $\bigg( 1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1} \bigg)^{\frac{1}{n}} > 1$ Letting $n \to \infty$ we have, $\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} > \lim_{n \to \infty}a_1 =a_1$ Thus by $Sandwhich-theorem$ $\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} = a_1$