find at least 10 partial sums of the series. (round your answers to five decimal places.)
graph both the sequence of terms and the sequence of partial sums on the same screen. does it appear that the series is convergent or divergent? if it is convergent, find the sum. (if the quantity diverges, enter diverges.)
if it is divergent, explain why.
o the terms of the series do not approach 0.
o the sequence of partial sums is divergent.
o the series is convergent.

Question

Accepted Answer

# Explanation:
## Step1: Recall the formula for the $n$ - th partial sum of a geometric series
The general form of a geometric series is $\sum_{n = 1}^{\infty}ar^{n - 1}$, and its $n$ - th partial sum is $S_{n}=\frac{a(1 - r^{n})}{1 - r}$ when $r
eq1$. For the series $\sum_{n = 1}^{\infty}\frac{1}{8(-3)^{n}}$, we can rewrite it as $\sum_{n = 1}^{\infty}\frac{1}{8}\times(-\frac{1}{3})^{n}$. Here, $a=\frac{1}{8}\times(-\frac{1}{3})$ and $r =-\frac{1}{3}$.
## Step2: Calculate the $n$ - th partial sum
Using the formula $S_{n}=\frac{a(1 - r^{n})}{1 - r}$, substitute $a =-\frac{1}{24}$ and $r=-\frac{1}{3}$ into the formula.
$$
\begin{align*}
S_{n}&=\frac{-\frac{1}{24}(1-(-\frac{1}{3})^{n})}{1-(-\frac{1}{3})}\
&=\frac{-\frac{1}{24}(1-(-\frac{1}{3})^{n})}{\frac{4}{3}}\
&=-\frac{1}{32}(1 - (-\frac{1}{3})^{n})
\end{align*}
$$
## Step3: Calculate $S_{n}$ for $n = 1$ to $n = 10$
- When $n = 1$:
$$
\begin{align*}
S_{1}&=-\frac{1}{32}(1-(-\frac{1}{3})^{1})\
&=-\frac{1}{32}\times\frac{4}{3}\
&\approx - 0.04167
\end{align*}
$$
- When $n = 2$:
$$
\begin{align*}
S_{2}&=-\frac{1}{32}(1-(-\frac{1}{3})^{2})\
&=-\frac{1}{32}\times\frac{8}{9}\
&\approx - 0.02778
\end{align*}
$$
- When $n = 3$:
$$
\begin{align*}
S_{3}&=-\frac{1}{32}(1-(-\frac{1}{3})^{3})\
&=-\frac{1}{32}\times\frac{28}{27}\
&\approx - 0.03241
\end{align*}
$$
- When $n = 4$:
$$
\begin{align*}
S_{4}&=-\frac{1}{32}(1-(-\frac{1}{3})^{4})\
&=-\frac{1}{32}\times\frac{80}{81}\
&\approx - 0.03086
\end{align*}
$$
- When $n = 5$:
$$
\begin{align*}
S_{5}&=-\frac{1}{32}(1-(-\frac{1}{3})^{5})\
&=-\frac{1}{32}\times\frac{244}{243}\
&\approx - 0.03175
\end{align*}
$$
- When $n = 6$:
$$
\begin{align*}
S_{6}&=-\frac{1}{32}(1-(-\frac{1}{3})^{6})\
&=-\frac{1}{32}\times\frac{730}{729}\
&\approx - 0.03149
\end{align*}
$$
- When $n = 7$:
$$
\begin{align*}
S_{7}&=-\frac{1}{32}(1-(-\frac{1}{3})^{7})\
&=-\frac{1}{32}\times\frac{2188}{2187}\
&\approx - 0.03161
\end{align*}
$$
- When $n = 8$:
$$
\begin{align*}
S_{8}&=-\frac{1}{32}(1-(-\frac{1}{3})^{8})\
&=-\frac{1}{32}\times\frac{6562}{6561}\
&\approx - 0.03155
\end{align*}
$$
- When $n = 9$:
$$
\begin{align*}
S_{9}&=-\frac{1}{32}(1-(-\frac{1}{3})^{9})\
&=-\frac{1}{32}\times\frac{19684}{19683}\
&\approx - 0.03158
\end{align*}
$$
- When $n = 10$:
$$
\begin{align*}
S_{10}&=-\frac{1}{32}(1-(-\frac{1}{3})^{10})\
&=-\frac{1}{32}\times\frac{59048}{59049}\
&\approx - 0.03156
\end{align*}
$$
## Step4: Determine convergence
Since $\vert r\vert=\vert-\frac{1}{3}\vert=\frac{1}{3}<1$, the series converges.
Using the formula for the sum of an infinite geometric series $S=\frac{a}{1 - r}$ (where $a =-\frac{1}{24}$ and $r =-\frac{1}{3}$), we have:
$$
\begin{align*}
S&=\frac{-\frac{1}{24}}{1-(-\frac{1}{3})}\
&=\frac{-\frac{1}{24}}{\frac{4}{3}}\
&=-\frac{1}{32}\approx - 0.03125
\end{align*}
$$

# Answer:
The partial sums for $n = 1$ to $n = 10$ are approximately:
- $n = 1$: $-0.04167$
- $n = 2$: $-0.02778$
- $n = 3$: $-0.03241$
- $n = 4$: $-0.03086$
- $n = 5$: $-0.03175$
- $n = 6$: $-0.03149$
- $n = 7$: $-0.03161$
- $n = 8$: $-0.03155$
- $n = 9$: $-0.03158$
- $n = 10$: $-0.03156$

The series is convergent and its sum is $-\frac{1}{32}\approx - 0.03125$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for the \(n\) - th partial sum of a geometric series

Step2: Calculate the \(n\) - th partial sum

Step3: Calculate \(S_{n}\) for \(n = 1\) to \(n = 10\)

Step4: Determine convergence

Answer: