Power Series Expansion for Logarithm of 1 + x

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Theorem

The Newton-Mercator series defines the natural logarithm function as a power series expansion:

\(\ds \map \ln {1 + x}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n\)
\(\ds \) \(=\) \(\ds x - \frac {x^2} 2 + \frac {x^3} 3 - \frac {x^4} 4 + \cdots\)

valid for all $x \in \R$ such that $-1 < x \le 1$.


Corollary

\(\ds \map \ln {1 - x}\) \(=\) \(\ds -\sum_{n \mathop = 1}^\infty \frac {x^n} n\)
\(\ds \) \(=\) \(\ds -x - \frac {x^2} 2 - \frac {x^3} 3 - \frac {x^4} 4 - \cdots\)

valid for $-1 \le x < 1$.


Proof

From Sum of Infinite Geometric Sequence, putting $-x$ for $x$:

$(1): \quad \ds \sum_{n \mathop = 0}^\infty \paren {-x}^n = \frac 1 {1 + x}$

for $-1 < x < 1$.


From Power Series Converges Uniformly within Radius of Convergence, $(1)$ is uniformly convergent on every closed interval within the interval $\openint {-1} 1$.

From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:

\(\ds \int_0^x \frac 1 {1 + t} \rd t\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \int_0^x \paren {-t}^n \rd t\)
\(\ds \leadsto \ \ \) \(\ds \map \ln {1 + x}\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{n + 1} } {n + 1}\) Primitive of Reciprocal and Integral of Power
\(\ds \leadsto \ \ \) \(\ds \map \ln {1 + x}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n\) letting $n \to n - 1$


This needs considerable tedious hard slog to complete it.
In particular: Cover the case where $x = 1$
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Sources

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