Question

test the series for convergence using the alternating series test.
sum _ { n = 1 } ^ { infty } 9 ( - 1 ) ^ { n } e ^ { - 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: 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 \(b_{n}\geq0\). Given the series \(\sum_{n=1}^{\infty}9(-1)^{n}e^{-n}\), we can rewrite it as \(\sum_{n = 1}^{\infty}(-1)^{n}\times9e^{-n}\).

Step2: Identify \(b_{n}\)

Comparing with the form \(\sum_{n = 1}^{\infty}(-1)^{n}b_{n}\), we get \(b_{n}=9e^{-n}\).

Step3: Check the conditions of the Alternating - Series Test

Condition 1: \(\lim_{n

ightarrow\infty}b_{n}\)
We know that \(y = e^{x}\) is an exponential function. Using the formula \(\lim_{n
ightarrow\infty}b_{n}=\lim_{n
ightarrow\infty}9e^{-n}\). Since \(\lim_{n
ightarrow\infty}e^{-n}=\lim_{n
ightarrow\infty}\frac{1}{e^{n}}\), and as \(n
ightarrow\infty\), \(e^{n}
ightarrow\infty\), so \(\lim_{n
ightarrow\infty}\frac{1}{e^{n}} = 0\). Then \(\lim_{n
ightarrow\infty}9e^{-n}=9\lim_{n
ightarrow\infty}e^{-n}=0\).

Condition 2: \(b_{n + 1}\leq b_{n}\) for all \(n\)

We have \(b_{n}=9e^{-n}\) and \(b_{n + 1}=9e^{-(n + 1)}\). Then \(\frac{b_{n+1}}{b_{n}}=\frac{9e^{-(n + 1)}}{9e^{-n}}=e^{-1}=\frac{1}{e}\lt1\). So \(b_{n + 1}=9e^{-(n + 1)}=\frac{9}{e^{n+1}}\lt\frac{9}{e^{n}} = b_{n}\) for all \(n\geq1\).

Answer:

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