Question

1. 2.28 / 2.85 points

test the series for convergence using the alternating series test.
sum _ { n = 1 } ^ { infty } ( - 1 ) ^ { n } \frac { 7 n - 5 } { 4 n + 5 }
identify ( b _ { n } ).
evaluate the following limit.
lim _ { n
ightarrow infty } b _ { n }
since ( lim _ { n
ightarrow infty } b _ { n } ) ( 0 ) and ( b _ { n + 1 } ) ( b _ { n } ) for all ( n ),

Explanation:

Step1: Recall the form of an alternating series

An alternating series is of the form \(\sum_{n = 1}^{\infty}(-1)^{n}b_{n}\) or \(\sum_{n = 1}^{\infty}(-1)^{n+1}b_{n}\), where \(b_{n}\geq0\) for all \(n\). For the series \(\sum_{n = 1}^{\infty}(-1)^{n}\frac{7n - 5}{4n+5}\), we can identify \(b_{n}=\frac{7n - 5}{4n + 5}\).

Step2: Evaluate the limit of \(b_{n}\) as \(n\to\infty\)

We use the formula for the limit of a rational function \(\lim_{n
ightarrow\infty}\frac{an + b}{cn + d}=\frac{a}{c}\) (when \(a
eq0\) and \(c
eq0\)).
For \(b_{n}=\frac{7n - 5}{4n + 5}\), we have \(\lim_{n
ightarrow\infty}b_{n}=\lim_{n
ightarrow\infty}\frac{7n-5}{4n + 5}\).
Divide both the numerator and denominator by \(n\):

$$
LATEXBLOCK0
$$

Since \(\lim_{n
ightarrow\infty}\frac{1}{n}=0\), we get \(\lim_{n
ightarrow\infty}\frac{7-\frac{5}{n}}{4+\frac{5}{n}}=\frac{7}{4}\)

Step3: Check the conditions of the Alternating - Series Test

The Alternating - Series Test states that for a series \(\sum_{n = 1}^{\infty}(-1)^{n}b_{n}\), if \(\lim_{n
ightarrow\infty}b_{n}=0\) and \(b_{n + 1}\leq b_{n}\) for all \(n\) (eventually), then the series converges.
Here, \(\lim_{n
ightarrow\infty}b_{n}=\frac{7}{4}
eq0\).

Answer:

Since \(\lim_{n
ightarrow\infty}b_{n}=\frac{7}{4}
eq0\), the series \(\sum_{n = 1}^{\infty}(-1)^{n}\frac{7n - 5}{4n+5}\) diverges by the Alternating - Series Test (because one of the necessary conditions \(\lim_{n
ightarrow\infty}b_{n}=0\) is not met).