Question

1. 1.16 / 1.92 pointstest the series for convergence using the alternating series test.$$ \frac { 1 } { \ln ( 2 ) } - \frac { 1 } { \ln ( 3 ) } + \frac { 1 } { \ln ( 4 ) } - \frac { 1 } { \ln ( 5 ) } + \frac { 1 } { \ln ( 6 ) } - \dots $$identify $ b _ { n } $. (assume the series starts at $ n = 1 $.)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 $, the series converges

Explanation:

Step1: Identify \(b_n\)

For an alternating series \(\sum_{n = 1}^{\infty}(- 1)^{n + 1}b_n\) (or \(\sum_{n=1}^{\infty}(-1)^{n - 1}b_n\)), given the series \(\frac{1}{\ln(2)}-\frac{1}{\ln(3)}+\frac{1}{\ln(4)}-\frac{1}{\ln(5)}+\frac{1}{\ln(6)}-\cdots\), we can rewrite it in the form \(\sum_{n = 1}^{\infty}(-1)^{n + 1}\frac{1}{\ln(n + 1)}\). So, \(b_n=\frac{1}{\ln(n + 1)}\).

Step2: Evaluate \(\lim_{n

ightarrow\infty}b_n\)
We know that \(\lim_{n
ightarrow\infty}b_n=\lim_{n
ightarrow\infty}\frac{1}{\ln(n + 1)}\). As \(n
ightarrow\infty\), \(n+1
ightarrow\infty\). And we know the property of the natural - logarithm function: \(\lim_{x
ightarrow\infty}\ln(x)=\infty\). Using the quotient rule for limits \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}\) (when \(\lim_{x
ightarrow a}g(x)
eq0\)), here \(f(x) = 1\) and \(g(x)=\ln(x + 1)\), \(\lim_{n
ightarrow\infty}\frac{1}{\ln(n + 1)}=\frac{\lim_{n
ightarrow\infty}1}{\lim_{n
ightarrow\infty}\ln(n + 1)}=\frac{1}{\infty}=0\).

Answer:

\(b_n=\frac{1}{\ln(n + 1)}\), \(\lim_{n
ightarrow\infty}b_n = 0\)