北京信息科技大学促进高校内涵发展-学科建设类项目-一级学科—石英惯性传感技术及器件研究开发平台建设合同公告

021

2023-05-25

Find the sum of the series $1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots$.

qwen2021

The series given is a geometric series of the form:
\[ 1 + \frac{1}{3} + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 + \cdots \]
where the first term \( a = 1 \) and the common ratio \( r = \frac{1}{3} \).

The sum \( S \) of an infinite geometric series with \( |r| < 1 \) is given by the formula:
\[ S = \frac{a}{1 - r} \]

Substituting the values \( a = 1 \) and \( r = \frac{1}{3} \) into the formula, we get:
\[ S = \frac{1}{1 - \frac{1}{3}} \]

First, simplify the denominator:
\[ 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3} \]

Now, substitute back into the sum formula:
\[ S = \frac{1}{\frac{2}{3}} = \frac{1 \cdot 3}{2} = \frac{3}{2} \]

Therefore, the sum of the series is:
\[ \boxed{\frac{3}{2}} \]

The answer is \(\boxed{\frac{3}{2}}\).

qwen2021

2023-05-01

Find the sum of the series $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$

user994715

The given series is:

\[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \]

This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = \frac{1}{2} \).

The sum \( S \) of an infinite geometric series with \( |r| < 1 \) is given by the formula:

\[ S = \frac{a}{1 - r} \]

Substituting the values \( a = 1 \) and \( r = \frac{1}{2} \) into the formula, we get:

\[ S = \frac{1}{1 - \frac{1}{2}} \]

First, simplify the denominator:

\[ 1 - \frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2} \]

Now, substitute back into the sum formula:

\[ S = \frac{1}{\frac{1}{2}} = 1 \times 2 = 2 \]

Therefore, the sum of the series is:

\[ \boxed{2} \]

qwen2021

2022-11-06

How do you find the sum of the infinite geometric series $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$?

user994715

To find the sum of the infinite geometric series \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\), we will use the formula for the sum of an infinite geometric series. The formula is given by:

\[ S = \frac{a}{1 - r} \]

where:

• \( a \) is the first term of the series.


• \( r \) is the common ratio.

Step-by-Step Solution:

1. Identify the first term \( a \):
The first term \( a \) is the value of the series when \( n = 0 \). In this series, the first term is \( a = 1 \).

2. Identify the common ratio \( r \):
The common ratio \( r \) is the factor by which each term is multiplied to get the next term. In this series, each term is half of the previous term:
\[
\frac{\frac{1}{2}}{1} = \frac{1}{2}, \quad \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}, \quad \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2}, \quad \text{and so on.}
\]
Therefore, the common ratio \( r = \frac{1}{2} \).

3. Apply the formula for the sum of an infinite geometric series:
Since \( |r| < 1 \), the series converges, and we can use the formula:
\[
S = \frac{a}{1 - r}
\]
Substituting the values \( a = 1 \) and \( r = \frac{1}{2} \):
\[
S = \frac{1}{1 - \frac{1}{2}}
\]

4. Simplify the expression:
\[
S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{