Question

test the series for convergence using the alternating series test.
-\frac{2}{4}+\frac{4}{5}-\frac{6}{6}+\frac{8}{7}-\frac{10}{8}+\cdots
identify ( b_{n} ). (assume the series starts at ( n = 1 ).)
evaluate the following limit.
( lim _{n-infty} b_{n} )
since ( lim _{n-infty} b_{n}
eq 0 ) and ( b_{n+1} geq b_{n} ) for all ( n ), the test is inconclusive.

Explanation:

Step1: Identify the general term of the alternating series

For an alternating series \(\sum_{n = 1}^{\infty}(- 1)^{n}b_{n}\), in the given series \(-\frac{2}{4}+\frac{4}{5}-\frac{6}{6}+\frac{8}{7}-\frac{10}{8}+\cdots\), we can rewrite it as \(\sum_{n = 1}^{\infty}(-1)^{n}\frac{2n}{n + 3}\). So \(b_{n}=\frac{2n}{n + 3}\).

Step2: Evaluate the limit \(\lim_{n

ightarrow\infty}b_{n}\)
We use the formula for the limit of a rational function. Divide both the numerator and denominator of \(b_{n}=\frac{2n}{n + 3}\) by \(n\). Then \(b_{n}=\frac{2}{1+\frac{3}{n}}\).
As \(n
ightarrow\infty\), \(\lim_{n
ightarrow\infty}\frac{3}{n}=0\). So \(\lim_{n
ightarrow\infty}b_{n}=\lim_{n
ightarrow\infty}\frac{2}{1 + \frac{3}{n}}=2\).

Answer:

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