Question

test the series for convergence using the alternating series test.
sum _ { n = 1 } ^ { infty } ( - 1 ) ^ { n } \frac { n ^ { 2 } } { n ^ { 3 } + 6 }
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 > 3 ),

Explanation:

Step1: Identify \(b_n\)

For an alternating series \(\sum_{n = 1}^{\infty}(-1)^{n}a_{n}\), in the given series \(\sum_{n = 1}^{\infty}(-1)^{n}\frac{n^{2}}{n^{3}+6}\), we have \(b_{n}=\frac{n^{2}}{n^{3}+6}\).

Step2: Evaluate \(\lim_{n

ightarrow\infty}b_{n}\)
Divide numerator and denominator by \(n^{3}\):

$$
LATEXBLOCK0
$$

As \(n
ightarrow\infty\), \(\frac{1}{n}
ightarrow0\) and \(\frac{6}{n^{3}}
ightarrow0\). So \(\lim_{n
ightarrow\infty}b_{n}=0\).

Answer:

\(b_{n}=\frac{n^{2}}{n^{3}+6}\), \(\lim_{n
ightarrow\infty}b_{n} = 0\)