Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 9 - Infinite Series - 9.1 Sequences - Exercises Set 9.1 - Page 606: 23


General term: $$\bigg\{ \frac{2n-1}{2n} \bigg\}_{n=1}^{+\infty}$$ The sequence converges to $2$.

Work Step by Step

As can be seen from the question, the numerators are odd numbers whereas the denominators are all even numbers. So, $n^{th}$ odd number is given by $(2n-1)$ and $n^{th}$ even number is given by $(2n)$. Thus, general term is $$\frac{2n-1}{2n} $$ The limit $$\lim_{n\to\infty}{ \frac{2n-1}{2n}}=1$$ because degree of n in both numerator and denominator is same, i.e., $1$.
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