Math / Calculus

QuestionFill in each blank. $\begin{array}{l} 6+\frac{6}{5}+\frac{6}{25}+\frac{6}{125} \ldots \\ \lim _{n \rightarrow \infty} a_{n}= \\ \lim _{n \rightarrow \infty} S_{n}= \end{array}$ $\square$ $\square$

Studdy Solution

STEP 1

1. The sequence given is a geometric series.
2. The general form of a geometric series is $a + ar + ar^2 + ar^3 + \ldots$ .
3. The sum of an infinite geometric series $S$ is given by $S = \frac{a}{1-r}$ , where $|r| < 1$ .

STEP 2

1. Identify the first term $a$ and the common ratio $r$ of the series.
2. Determine the limit of the sequence term $a_n$ as $n \rightarrow \infty$ .
3. Determine the sum of the infinite series $S_n$ .

STEP 3

Identify the first term $a$ and the common ratio $r$ of the series:
The series is $6 + \frac{6}{5} + \frac{6}{25} + \frac{6}{125} + \ldots$ .
The first term $a = 6$ .
The common ratio $r$ is found by dividing the second term by the first term:
$r = \frac{\frac{6}{5}}{6} = \frac{1}{5}$

STEP 4

Determine the limit of the sequence term $a_n$ as $n \rightarrow \infty$ :
The general term of the sequence $a_n = 6 \left(\frac{1}{5}\right)^{n-1}$ .
As $n \rightarrow \infty$ , the term $\left(\frac{1}{5}\right)^{n-1} \rightarrow 0$ .
Thus, $\lim_{n \rightarrow \infty} a_n = 0$ .

STEP 5

Determine the sum of the infinite series $S_n$ :
The sum of an infinite geometric series is given by:
$S = \frac{a}{1-r} = \frac{6}{1-\frac{1}{5}} = \frac{6}{\frac{4}{5}} = 6 \times \frac{5}{4} = \frac{30}{4} = 7.5$
Thus, $\lim_{n \rightarrow \infty} S_n = 7.5$ .
The completed blanks are:
1. $\lim_{n \rightarrow \infty} a_n = 0$
2. $\lim_{n \rightarrow \infty} S_n = 7.5$

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