Power Series Expansion for Logarithm of 1 + x over 1 + x
Jump to navigation
Jump to search
Theorem
\(\ds \frac {\map \ln {1 + x} } {1 + x}\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n + 1} H_n x^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x - H_2 x^2 + H_3 x^3 - H_4 x^4 + \cdots\) |
where $H_n$ denotes the $n$th harmonic number:
- $H_n = \ds \sum_{r \mathop = 1}^n \dfrac 1 r = 1 + \dfrac 1 2 + \dfrac 1 3 \cdots + \dfrac 1 r$
valid for all $x \in \R$ such that $\size x < 1$.
Proof
Let $\map f x = \dfrac {\map \ln {1 + x} } {1 + x}$.
By definition of Maclaurin series:
- $(1): \quad \map f x \sim \ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!} \map {f^{\paren n} } 0$
where $\map {f^{\paren n} } 0$ denotes the $n$th derivative of $f$ with respect to $x$ evaluated at $x = 0$.
From Nth Derivative of Natural Logarithm by Reciprocal:
- $\dfrac {\d^n} {\d x^n} \dfrac {\map \ln {1 + x} } {1 + x} = \paren {-1}^{n + 1} n! \dfrac {H_n - \map \ln {1 + x} } {\paren {1 + x}^{n + 1} }$
The result follows by setting $x = 0$ and substituting for $\map {f^{\paren n} } 0$ in $(1)$.
This needs considerable tedious hard slog to complete it. In particular: Prove the range of convergence To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 20$: Miscellaneous Series: $20.51$