Question

1. 2.28 / 2.85 points

test the series for convergence using the alternating series test.
$$ sum _ { n = 1 } ^ { infty } \frac { ( - 1 ) ^ { n - 1 } } { 4 + 9 n } $$
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: Identify \(b_n\)

For an alternating series \(\sum_{n = 1}^{\infty}(-1)^{n - 1}a_n\), in the given series \(\sum_{n=1}^{\infty}\frac{(-1)^{n - 1}}{4 + 9n}\), we can identify \(b_n=\frac{1}{4 + 9n}\) according to the form of alternating series \(a_n = b_n\) (where the alternating part is \((-1)^{n-1}\)).

Step2: Evaluate \(\lim_{n

ightarrow\infty}b_n\)
We use the limit formula for rational functions. If \(b_n=\frac{1}{4 + 9n}\), then \(\lim_{n
ightarrow\infty}b_n=\lim_{n
ightarrow\infty}\frac{1}{4 + 9n}\).
As \(n
ightarrow\infty\), the denominator \(4 + 9n
ightarrow\infty\). So, \(\lim_{n
ightarrow\infty}\frac{1}{4 + 9n}=0\) (using the rule \(\lim_{n
ightarrow\infty}\frac{c}{an + b}=0\) for \(a>0\) and \(c,b\) constants).

Step3: Check \(b_{n+1}\leq b_n\)

We have \(b_n=\frac{1}{4 + 9n}\) and \(b_{n + 1}=\frac{1}{4+9(n + 1)}=\frac{1}{13 + 9n}\).
Since \(4 + 9n<13 + 9n\) for \(n\geq1\), then \(\frac{1}{4 + 9n}>\frac{1}{13 + 9n}\), i.e., \(b_{n+1}\leq b_n\) for all \(n\geq1\).

Answer:

\(b_n=\frac{1}{4 + 9n}\), \(\lim_{n
ightarrow\infty}b_n = 0\), and since \(\lim_{n
ightarrow\infty}b_n=0\) and \(b_{n + 1}\leq b_n\) for all \(n\), the series \(\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{4+9n}\) converges.