Kuzee
Infinite Series

Logarithmic Series

Definition of Logarithmic Series

Definition of Logarithmic Series

The series ∑n=1∞(−1)n+1nxn\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n is called the logarithmic series.

SymbolTypePronunciation/ExplanationMeaning in This Article
∑\sumGreek letterSigmaSummation symbol, representing series
∞\inftyMathematical symbolInfinityRepresents infinite series, infinite number of terms
ln⁡\lnMathematical symbolNatural logarithmNatural logarithm function, ln⁡x=log⁡ex\ln x = \log_e x

Convergence

Convergence of Logarithmic Series
  • When ∣x∣<1|x| < 1, the series converges
  • When x=1x = 1, the series converges (Leibniz test)
  • The sum is:

∑n=1∞(−1)n+1nxn=ln⁡(1+x)\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n = \ln(1+x)

Proof

When ∣x∣<1|x| < 1

Use the ratio test:

an+1an=(−1)n+2n+1xn+1⋅n(−1)n+1xn=−nn+1x\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+2}}{n+1} x^{n+1} \cdot \frac{n}{(-1)^{n+1} x^n} = -\frac{n}{n+1} x

lim⁡n→∞∣an+1an∣=lim⁡n→∞nn+1∣x∣=∣x∣<1\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{n}{n+1} |x| = |x| < 1

So the series converges.

When x=1x = 1

The series becomes ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}, which is an alternating series, satisfying the conditions of the Leibniz test, so it also converges.

Examples

Example 1

Find the sum of the series ∑n=1∞(−1)n+1n(12)n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{1}{2}\right)^n.

Solution: This is a logarithmic series with x=12x = \frac{1}{2}

So the sum is: ln⁡(1+12)=ln⁡32\ln\left(1 + \frac{1}{2}\right) = \ln\frac{3}{2}

Example 2

Find the sum of the series ∑n=1∞(−1)n+1n(13)n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{1}{3}\right)^n.

Solution: This is a logarithmic series with x=13x = \frac{1}{3}

So the sum is: ln⁡(1+13)=ln⁡43\ln\left(1 + \frac{1}{3}\right) = \ln\frac{4}{3}

Example 3

Find the sum of the series ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}.

Solution: This is a logarithmic series with x=1x = 1

So the sum is: ln⁡(1+1)=ln⁡2\ln(1 + 1) = \ln 2

Exercises

Exercise 1

Find the sum of the series ∑n=1∞(−1)n+1n(14)n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{1}{4}\right)^n.

logarithmic seriesseries summation

Approach: This is a logarithmic series; identify the value of xx.

Steps:

  1. Identify the series type: ∑n=1∞(−1)n+1n(14)n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{1}{4}\right)^n is a logarithmic series
  2. Determine the parameter: x=14x = \frac{1}{4}
  3. Compute the sum: S=ln⁡(1+14)=ln⁡54S = \ln\left(1 + \frac{1}{4}\right) = \ln\frac{5}{4}

Answer: The sum is ln⁡54\ln\frac{5}{4}.

Exercise 2

Find the sum of the series ∑n=1∞(−1)n+1n(−12)n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(-\frac{1}{2}\right)^n.

logarithmic seriesseries summation

Approach: This is a logarithmic series; identify the value of xx.

Steps:

  1. Identify the series type: ∑n=1∞(−1)n+1n(−12)n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(-\frac{1}{2}\right)^n is a logarithmic series
  2. Determine the parameter: x=−12x = -\frac{1}{2}
  3. Compute the sum: S=ln⁡(1−12)=ln⁡12=−ln⁡2S = \ln\left(1 - \frac{1}{2}\right) = \ln\frac{1}{2} = -\ln 2

Answer: The sum is −ln⁡2-\ln 2.

Exercise 3

Find the sum of the series ∑n=1∞(−1)n+1n(23)n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{2}{3}\right)^n.

logarithmic seriesseries summation

Approach: This is a logarithmic series; identify the value of xx.

Steps:

  1. Identify the series type: ∑n=1∞(−1)n+1n(23)n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{2}{3}\right)^n is a logarithmic series
  2. Determine the parameter: x=23x = \frac{2}{3}
  3. Compute the sum: S=ln⁡(1+23)=ln⁡53S = \ln\left(1 + \frac{2}{3}\right) = \ln\frac{5}{3}

Answer: The sum is ln⁡53\ln\frac{5}{3}.

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
xxMathematical symbolVariableVariable in logarithmic series
lim⁡\limMathematical symbolLimitRepresents limit of sequence or function
Previous
Exponential Series
Next
Power Series

Found an issue with this content?

Tell us what could be improved.