Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 30


This series diverges.

Work Step by Step

Using the test for divergence that is stated in page 753 of the textbook we will take the limit of the corresponding sequence to the given series. Meaning, we will solve: $\lim\limits_{n \to \infty} \frac{k^{2}}{k^{2}-2k+5}$, to solve this we take the highest power of $k$ from the numerator and the denominator. In this case, we factor out $k^{2}$ and the result looks like: $\lim\limits_{n \to \infty} \frac{1}{1-\frac{2}{k}-\frac{5}{k^{2}} }$ and taking $k$ towards $\infty$, the fractions with $k$ at their denominator become zero and what remains is: $1$ $\ne$ 0 . Thus by test for divergence our series diverges.
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