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 $Sandwhich-theorem$ \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} = a_1\)