Question

determine whether the series is convergent or divergent. if it is convergent, find its sum. (if the quantity diverges, enter diverges.)
$$ sum _ { n = 1 } ^ { infty } \frac { 1 } { 8 + e ^ { - n } } $$

Explanation:

Step1: Identify the series type

This is a geometric series of the form \(\sum_{n = 1}^{\infty}ar^{n - 1}\). Rewrite \(\sum_{n=1}^{\infty}\frac{1}{8 + e^{\frac{n}{4}}}\). Let's check the limit of the general term \(a_{n}=\frac{1}{8 + e^{\frac{n}{4}}}\) as \(n
ightarrow\infty\).
We know that \(\lim_{n
ightarrow\infty}e^{\frac{n}{4}}=\infty\) (since the exponential function \(y = e^{x}\), when \(x=\frac{n}{4}\) and \(n
ightarrow\infty\), \(x
ightarrow\infty\)). Then \(\lim_{n
ightarrow\infty}a_{n}=\lim_{n
ightarrow\infty}\frac{1}{8 + e^{\frac{n}{4}}}\).
Using the rule \(\lim_{x
ightarrow\infty}\frac{c}{x}=0\) (\(c\) is a constant) and the properties of limits \(\lim_{n
ightarrow\infty}\frac{1}{8 + e^{\frac{n}{4}}}=\frac{\lim_{n
ightarrow\infty}1}{\lim_{n
ightarrow\infty}(8 + e^{\frac{n}{4}})}=\frac{1}{\infty}=0\). But we can also use the comparison test.
We know that \(e^{\frac{n}{4}}>0\) for \(n\geq1\), so \(8 + e^{\frac{n}{4}}>e^{\frac{n}{4}}\). Then \(\frac{1}{8 + e^{\frac{n}{4}}}<\frac{1}{e^{\frac{n}{4}}}=e^{-\frac{n}{4}}\).

Step2: Apply the comparison test

The series \(\sum_{n = 1}^{\infty}e^{-\frac{n}{4}}\) is a geometric series with \(a = e^{-\frac{1}{4}}\) and \(r=e^{-\frac{1}{4}}\). For a geometric series \(\sum_{n=1}^{\infty}ar^{n - 1}\), the sum is \(S=\frac{a}{1 - r}\) when \(\vert r\vert<1\). Here, \(\vert r\vert=\vert e^{-\frac{1}{4}}\vert=\frac{1}{e^{\frac{1}{4}}}<1\) (since \(e^{\frac{1}{4}}>1\)).
The sum of \(\sum_{n = 1}^{\infty}e^{-\frac{n}{4}}\) is \(S=\frac{e^{-\frac{1}{4}}}{1 - e^{-\frac{1}{4}}}=\frac{1}{e^{\frac{1}{4}}- 1}\).
By the comparison test, if \(0\leq a_{n}\leq b_{n}\) for all \(n\geq N\) (in this case \(N = 1\)) and \(\sum_{n = 1}^{\infty}b_{n}\) converges, then \(\sum_{n = 1}^{\infty}a_{n}\) converges.

Answer:

The series \(\sum_{n = 1}^{\infty}\frac{1}{8 + e^{\frac{n}{4}}}\) is convergent.