Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Section 13.4 - Limits at Infinity; Limits of Sequences - 13.4 Exercises - Page 930: 31

Answer

Convergent with limit $\dfrac{3}{2}$

Work Step by Step

Given: $f(x)=\lim\limits_{n\to \infty} \dfrac{3}{n^2}[\dfrac{n(n+1)}{2}]=\lim\limits_{n\to \infty} \dfrac{3n^2+3n}{2n^2}$ The sequence converges when the limit $\lim\limits_{n\to \infty} a_n$ exists and when the limit $\lim\limits_{n\to \infty} a_n$ does not exist, then sequence diverges. Here, we have $\lim\limits_{n\to \infty}a_n=\lim\limits_{n\to \infty}\dfrac{3n^2+3n}{2n^2}$ This gives: $\lim\limits_{n\to \infty}a_n=\lim\limits_{n\to \infty}\dfrac{3+3/n}{2}$ or, $=\dfrac{\lim\limits_{n\to \infty}(3)+\lim\limits_{n\to \infty}(3/n)}{\lim\limits_{n\to \infty} (2)}$ Thus, $a_n=\dfrac{3}{2}$ Hence, the sequence is convergent with limit $\dfrac{3}{2}$
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