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

[donutmax]

hi!

are the following power series equivalent?

ln(1+x)=$$\sum_{n=0}^{\infty} \frac{(-1)^n n! x^{n+1}}{(n+1)!}$$
=$$\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}$$

 

[The Chaz]

Yes.

 

[CRGreathouse]

$$\frac{(-1)^n n! x^{n+1}}{(n+1)!}=\frac{(-1)^n x^{n+1}}{(n+1)!/n!}=\frac{(-1)^n x^{n+1}}{(n+1)}$$

 

[The Chaz]

$$\frac{(-1)^n n! x^{n+1}}{(n+1)!}=\frac{(-1)^n x^{n+1}}{(n+1)!/n!}=\frac{(-1)^n x^{n+1}}{(n+1)}$$

He's everywhere I want to be!

 

[CellCoree]

Yes, the two power series are equivalent. This is because ln(1+x) can be represented as the natural logarithm of (1+x), which can be expanded using the Maclaurin series. The Maclaurin series for ln(1+x) is given by:

ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...

This series can be written in summation form as:

\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}

which is equivalent to the second power series given in the question. Therefore, ln(1+x) can indeed be represented as a power series.