Can I Represent ln(1+x) as a Power Series?

AI Thread Summary
The discussion confirms that the power series representation of ln(1+x) can be expressed in two equivalent forms. The first form is ln(1+x) = ∑ (from n=0 to ∞) [(-1)^n n! x^(n+1) / (n+1)!], while the second is ln(1+x) = ∑ (from n=0 to ∞) [(-1)^n x^(n+1) / (n+1)]. The equivalence is demonstrated through algebraic manipulation, showing that both series yield the same result. The simplification reveals that the factorial terms cancel appropriately, confirming their equivalence. This validates the power series representation of the natural logarithm function.
donutmax
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hi!

are the following power series equivalent?

ln(1+x)=\sum_{n=0}^{\infty} \frac{(-1)^n n! x^{n+1}}{(n+1)!}
=\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}
 
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Yes.
 
\frac{(-1)^n n! x^{n+1}}{(n+1)!}=\frac{(-1)^n x^{n+1}}{(n+1)!/n!}=\frac{(-1)^n x^{n+1}}{(n+1)}
 
CRGreathouse said:
\frac{(-1)^n n! x^{n+1}}{(n+1)!}=\frac{(-1)^n x^{n+1}}{(n+1)!/n!}=\frac{(-1)^n x^{n+1}}{(n+1)}

He's everywhere I want to be!
 
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