Math Problem Statement

for the series ((1)/(10))^(n) --> infinity to n=1) what is first give partial sums of this series, what is the first term, the ratio and the sum of th ewhole infinite series?

Solution

The series you have given is a geometric series:

S=n=1(110)nS = \sum_{n=1}^{\infty} \left( \frac{1}{10} \right)^n

This series is a geometric series, where the general form is:

S=a+ar+ar2+ar3+S = a + ar + ar^2 + ar^3 + \dots

Where:

  • aa is the first term.
  • rr is the common ratio.

First Term:

From the given series, the first term (aa) is:

a=(110)1=110a = \left( \frac{1}{10} \right)^1 = \frac{1}{10}

Common Ratio:

The common ratio (rr) is the constant factor between successive terms, which is also:

r=110r = \frac{1}{10}

Sum of the Infinite Series:

For a geometric series, the sum of an infinite series is given by the formula:

S=a1rS_{\infty} = \frac{a}{1 - r}

This formula works when r<1|r| < 1. In your case, r=110r = \frac{1}{10}, which satisfies this condition, so we can use the formula:

S=1101110=110910=19S_{\infty} = \frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{\frac{1}{10}}{\frac{9}{10}} = \frac{1}{9}

Thus, the sum of the infinite series is:

S=19S_{\infty} = \frac{1}{9}

Partial Sums:

To calculate the partial sums of this series, we use the formula for the sum of the first nn terms of a geometric series:

Sn=a1rn1rS_n = a \frac{1 - r^n}{1 - r}

Substituting the values of a=110a = \frac{1}{10} and r=110r = \frac{1}{10}:

Sn=1101(110)n1110=1101(110)n910=1(110)n9S_n = \frac{1}{10} \cdot \frac{1 - \left( \frac{1}{10} \right)^n}{1 - \frac{1}{10}} = \frac{1}{10} \cdot \frac{1 - \left( \frac{1}{10} \right)^n}{\frac{9}{10}} = \frac{1 - \left( \frac{1}{10} \right)^n}{9}

Example of First Few Partial Sums:

  • For n=1n = 1, the partial sum is:

S1=1(110)19=11109=9109=110S_1 = \frac{1 - \left( \frac{1}{10} \right)^1}{9} = \frac{1 - \frac{1}{10}}{9} = \frac{\frac{9}{10}}{9} = \frac{1}{10}

  • For n=2n = 2, the partial sum is:

S2=1(110)29=111009=991009=11100S_2 = \frac{1 - \left( \frac{1}{10} \right)^2}{9} = \frac{1 - \frac{1}{100}}{9} = \frac{\frac{99}{100}}{9} = \frac{11}{100}

  • For n=3n = 3, the partial sum is:

S3=1(110)39=1110009=99910009=1111000S_3 = \frac{1 - \left( \frac{1}{10} \right)^3}{9} = \frac{1 - \frac{1}{1000}}{9} = \frac{\frac{999}{1000}}{9} = \frac{111}{1000}

Summary:

  • First term (a): 110\frac{1}{10}
  • Common ratio (r): 110\frac{1}{10}
  • Sum of the infinite series: 19\frac{1}{9}
  • Partial sums:
    • S1=110S_1 = \frac{1}{10}
    • S2=11100S_2 = \frac{11}{100}
    • S3=1111000S_3 = \frac{111}{1000}

Do you want to explore more partial sums or have any questions about the process?


Further Questions:

  1. What is the behavior of the partial sums as nn increases?
  2. How do geometric series differ from arithmetic series?
  3. What is the convergence criterion for a geometric series?
  4. How does the sum of a geometric series change if r>1r > 1?
  5. How would you derive the sum of a geometric series with negative terms?

Tip: The convergence of a geometric series is highly dependent on the common ratio rr. If r<1|r| < 1, the series will converge; otherwise, it will diverge.

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Math Problem Analysis

Mathematical Concepts

Geometric Series
Summation of Infinite Series
Partial Sums

Formulas

Sum of Infinite Geometric Series: S = a / (1 - r)
Partial Sum of Geometric Series: S_n = a * (1 - r^n) / (1 - r)

Theorems

Geometric Series Convergence Criterion

Suitable Grade Level

Grades 9-12