Simplifying and Proving the Series: ln(1 e^x e^{2x} e^{3x} ...) -ln(1 - e^x)

The mathematical problem discussed below aims to prove that the natural logarithm of the infinite series 1 e^x e^{2x} e^{3x} ... equals -ln(1 - e^x). This proof involves recognizing the series as a geometric series, summing the series, and then applying logarithmic properties.

Introduction

The series in question, 1 e^x e^{2x} e^{3x} ..., is an infinite geometric series with the first term a 1 and the common ratio r e^x. The sum of an infinite geometric series is given by:

S frac{a}{1 - r}

In our case, the sum is:

S frac{1}{1 - e^x}

where the series converges when |e^x|

Step 1: Summing the Geometric Series

The series 1 e^x e^{2x} e^{3x} ... can be expressed as:

1 frac{1(1)(e^x)}{1 - e^x}

Combining the terms, we get:

1 frac{e^x}{1 - e^x} frac{(1 - e^x) e^x}{1 - e^x} frac{1}{1 - e^x}

Step 2: Applying the Natural Logarithm

To prove the desired result, we take the natural logarithm of both sides of the equation:

ln(1 e^x e^{2x} e^{3x} ...) ln(1/1 - e^x)

Using the property of logarithms that ln(a/b) ln(a) - ln(b), we have:

ln(1 - e^x) ln(1) - ln(1 - e^x)

Since ln(1) 0, the equation simplifies to:

ln(1 - e^x) -ln(1 - e^x)

Conclusion

Therefore, the equality holds for x

ln(1 e^x e^{2x} e^{3x} ...) -ln(1 - e^x)

Summary and Key Points

The series 1 e^x e^{2x} e^{3x} ... is a geometric series converging when x The sum of the series is S frac{1}{1 - e^x}. By applying the natural logarithm, we prove that ln(1 e^x e^{2x} e^{3x} ...) -ln(1 - e^x).

Related Keywords

ln series, geometric series, logarithmic functions