Taylor series of log(1+z)

Theorem

The following formula holds for $|z| \leq 1$ and $z \neq -1$: $$\log(1+z)=-\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^kz^k}{k},$$ where $\log$ denotes the logarithm.

References

• 1964: {{ #if: |{{{2}}}|Milton Abramowitz}}{{#if: Irene A. Stegun|{{#if: |, {{ #if: |{{{2}}}|Irene A. Stegun}}{{#if: |, [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]{{#if: |, [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]}}| and [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]}}| and {{ #if: |{{{2}}}|Irene A. Stegun}}}}|}}: [[Book:Milton Abramowitz/Handbook of mathematical functions{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|Handbook of mathematical functions{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}}|{{#if: | ({{{ed}}} ed.)}}}}]]{{#if: | (translated by [[Mathematician:{{{translated}}}|{{ #if: |{{{2}}}|{{{translated}}}}}]])}}{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: Relationship between logarithm and logarithm base 10 | ... (previous)|}}{{#if: Series for log(z) for Re(z) greater than 1/2 | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}: 4.1.24