#### [SOLVED] What is the correct radius of convergence for $\ln(1+x)$?

My text tells me this:

And, Wolfram tells me this:

Now, I'm not certain what to believe, but I believe I'm not certain because I'm not certain if Wolfram is using the logarithm with base $10$. However, I believe that if Wolfram is correct then my text is wrong since its clear that at $x=1$ the series given by Wolfram diverges.

Which radius of convergence is correct?

#### @wythagoras 2015-07-10 15:46:12

Note that if $x=1$ then you get $$1-\frac12+\frac13-\frac14+\cdots$$

which is convergent. This can be seen by rewriting it using $$\frac1n-\frac1{n+1}=\frac1{n(n+1)}$$

$$\left(\frac1{2}\right)+\frac16+\frac1{30}+\frac1{56}+\cdots<\frac1{2}+\frac14+\frac1{25}+\frac1{49}+\cdots<\frac1{2}+\sum^{\infty}_{n=2}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{1}{2}$$

On the other hand, it is easy to see it is positive, furthermore, the differences $\frac{1}{k}$ are getting increasingly close to zero, so it will converge to a finite value.

#### @Mark Viola 2015-07-10 15:28:40

WA has an incomplete answer. Inasmuch as the McLaurin series for $\log (1+x)$ is

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

we see that the series converges for $|x|<1$ from, say for example, the ratio test. Checking the end points, we see that for $x=1$ the series is the convergent alternating harmonic series while for $x=-1$ the series is the divergent harmonic series. Thus, the interval of convergence is

$$-1<x\le 1$$

#### @Reinhild Van Rosenú 2015-07-10 15:31:50

Do you know a link that proves the alternating harmonic series converges.

#### @Mark Viola 2015-07-10 15:37:01

Yes. Here you will see a well-known general result. If the terms of an alternating series are decreasing to $0$, the series converges.

#### @Zain Patel 2015-07-10 15:27:05

Wolfram is using the natural logarithm, $\log x = \ln x$ in general, unless a base is specified.

And whilst Wolfram is correct, your textbook provides a more comprehensive reasoning, since at $x=1$, the series is the alternating harmonic series, which converges.

#### @Mark Viola 2015-07-10 15:29:50

At $x=1$, the series is the alternating harmonic series, which converges.

#### @Rory Daulton 2015-07-10 15:28:51

Both sources are correct, but your textbook is more comprehensive than Wolfram.

Any power series has a radius of convergence, where the series converges for any number inside the radius and diverges for any number outside the radius. Wolfram correctly says that the radius of convergence is $1$.

However, for real numbers, the two points at the radius of convergence may either converge or diverge. Wolfram does not address those points at all. Note that it says that the series converges for $|x|<1$: it does not say what happens when $|x|=1$. Your textbook does, and says it correctly.

The series does converge for $x=1$ since it is then an alternating series where the terms decrease in absolute value and tend to zero. I'm sure you know the theorem that says such a series converges. The series diverges at $x=-1$ since it is then the negative of the harmonic series, which famously diverges.