# 1. (a) Is the series sum_{n = 2}^{infty } {( - 1)^n} / {sqrt{n^2 - 1}} conditionally convergent,...

## Question:

1.

(a) Is the series {eq}\sum_{n=2}^{\infty }\frac{(-1)^{n}}{\sqrt{n^{2}-1}} {/eq} conditionally convergent, absolutely convergent or divergent? Justify your answer.

(b) Does the series {eq}\sum_{n=0}^{\infty }\frac{1}{2^{n}+\sqrt{n}} {/eq} converge or diverge?

(c) Does the series {eq}\sum_{n=1}^{\infty }\frac{\ln n}{\ln (n^{2}+1)} {/eq} converge or diverge?

(d) Determine whether the series {eq}\sum_{n=1}^{\infty }(-1)^{n}\frac{3^{n}}{n^{2}2^{n}} {/eq} converges absolutely, converges conditionally only or else diverges, justifying your answer.

## Absolute Convergence of a Series

Consider the series

{eq}S = \sum_{n=0}^\infty a_n. {/eq}

The series is said to be absolutely convergent if and only if

{eq}\sum_{n=0}^\infty |a_n| < \infty. {/eq}

A series that is absolutely convergent implies that the series is convergent. However the converse is not true as seen in the following example. Consider the alternating series,

{eq}S = \sum_{n=1}^\infty \frac{(-1)^n}{n} {/eq}

This is convergent due the Leibnitz' alternating series test. However it is not absolutely convergent since it is the harmonic series.

(a) Consider the series

{eq}S = \displaystyle \sum_{n=2}^{\infty }\frac{(-1)^{n}}{\sqrt{n^{2}-1}} {/eq}

Since the terms of the series alternate...

Become a Study.com member to unlock this answer! Create your account

Testing for Convergence & Divergence by Comparing Series

from AP Calculus BC: Exam Prep

Chapter 21 / Lesson 5
1.6K