Maclaurin series for natural log function

[tmt1]

I'm examining the Maclaurin series for $f(x) = ln(x + 1)$.

It is fairly straightforward but there are a few details I'm not getting.

So:

$$ ln(x + 1) = \int_{}^{} \frac{1}{1 + x}\,dx$$

which equals:

$A + x - \frac{x^2}{2}$ etc. or $A + \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{x^n}{n}$

I'm not sure what $A$ is supposed to be here. Is it the same as $C$, or a constant when evaluating the integral of a function?

The text says that because $ln(x + 1) = 0$ for $x = 0$ then $A $ equals $0$ and I'm not sure why this is the case.

Also, the text states that this is valid for $|x| < 1$ and for $x = 1$ and I'm not sure why this is the case.

 

[Prove It]

I'm examining the Maclaurin series for $f(x) = ln(x + 1)$.

It is fairly straightforward but there are a few details I'm not getting.

So:

$$ ln(x + 1) = \int_{}^{} \frac{1}{1 + x}\,dx$$

which equals:

$A + x - \frac{x^2}{2}$ etc. or $A + \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{x^n}{n}$

I'm not sure what $A$ is supposed to be here. Is it the same as $C$, or a constant when evaluating the integral of a function?

The text says that because $ln(x + 1) = 0$ for $x = 0$ then $A $ equals $0$ and I'm not sure why this is the case.

Also, the text states that this is valid for $|x| < 1$ and for $x = 1$ and I'm not sure why this is the case.

What happens when you substitute $\displaystyle \begin{align*} x = 0 \end{align*}$ into $\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \left( -1 \right) ^n \, \frac{x^n}{n} \end{align*}$?

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I'm examining the Maclaurin series for $f(x) = ln(x + 1)$.

It is fairly straightforward but there are a few details I'm not getting.

So:

$$ ln(x + 1) = \int_{}^{} \frac{1}{1 + x}\,dx$$

which equals:

$A + x - \frac{x^2}{2}$ etc. or $A + \sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{x^n}{n}$

I'm not sure what $A$ is supposed to be here. Is it the same as $C$, or a constant when evaluating the integral of a function?

The text says that because $ln(x + 1) = 0$ for $x = 0$ then $A $ equals $0$ and I'm not sure why this is the case.

Also, the text states that this is valid for $|x| < 1$ and for $x = 1$ and I'm not sure why this is the case.

I'm assuming you understand that the geometric series you started with has convergence where |x| < 1, and so that means your series you get through integrating also has the same interval of convergence except possibly at the endpoints.

Let $\displaystyle \begin{align*} x = 1 \end{align*}$ into $\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \left( -1 \right) ^n\,\frac{x^n}{n} \end{align*}$ and you get $\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \frac{\left( -1 \right) ^n }{n} \end{align*}$. As this is an alternating series, you can use the alternating series test to show this is convergent.

When x = -1, we have $\displaystyle \begin{align*} \sum_{n = 1}^{\infty} \left( -1 \right) ^n\,\frac{x^n}{n} = \sum_{n = 1}^{\infty} \left( -1 \right) ^n \, \frac{ \left( -1 \right) ^n }{n} = \sum_{n = 1}^{\infty} \frac{1}{n} \end{align*}$, which is the harmonic series, known to be divergent.