# The convergence of power series of $\log(1+x)$ with or without Taylor expansion

I would like to understand better the power series expansion of $$\log(1+x)$$ and how we know that it converges.

To begin, using the standard formula for Taylor expansion we quickly obtain:

$$\log(1+x) =-\sum_{k=1}^\infty\frac{(-x)^k}{k}~~.$$

Now, given that the RHS is a geometric series, it is clear that it will diverge for $$|x|>1$$. But then how do we know that for $$|x|<1$$, when the series does converge, that it converges to $$\log(1+x)$$?

I have found another "proof" that goes as follows. Let $$|x|<1$$ , then:

$$\log(1+x) =\int_0^x\frac{1}{1+t}~dt =\int_0^x\sum_{k=0}^\infty(-t)^k~dt =\sum_{k=0}^\infty\int_0^x(-t)^k~dt =-\sum_{k=1}^\infty\frac{(-x)^k}{k}~~.$$

We achieve the same result that is again valid for $$|x|<1$$, which is what we want. But we relied on term-wise integration and, as far as I know, the geometric series does not converge uniformly, so this step is not justified.

So to summarize my questions:

1. How can we justify that $$-\sum_{k=1}^\infty\frac{(-x)^k}{k}$$ does converge to $$\log(1+x)$$ for $$x\in(-1,1]$$?
2. How do we know, whether the Taylor sum of a function converges at all, and converges to the function?

Thanks a lot guys,

Alex

• For points $x$ in $(-1,1)$ you can do what you did with the integrals, since the convergence is uniform on $[-y,y]$ for any $x<y<1$. For $x=1$ you could use Littlewood's Tauberian theorem. Dec 7, 2019 at 13:15
• If the derivatives of the function in some interval grow (as a function of the order) not-faster than the terms of a power series, then you get convergence of the series to the function. You can see this from the form of the remainder in Lagrange form. This test is not so convenient, since it requires global bounds on the derivatives, but well, you can use it for some functions. It works for this one, for $\sin(x)$, $\cos(x)$, $e^x$ is easy to check too. Dec 7, 2019 at 13:19
• Above, where I said 'then the terms of a power series', I meant to say 'than the term of a geometric series'. Dec 7, 2019 at 13:28
• Ok, for some reason I thought that the geometric series wasn't uniformly convergent anywhere. Not sure why. Thanks, guys!
– Alex
Dec 7, 2019 at 14:44
• Note that one has uniform convergence on each compact set contained in the interval/disc of convergence, but this doesn't imply uniform convergence on the entire interval at once. Dec 7, 2019 at 14:46

The series $$\sum_{n=0}^\infty(-1)^nx^n$$ converges uniformly on reach interval $$[-r,r]$$, with $$r\in(0,1)$$. This is enough to justify the equality$$\int_0^x\sum_{n=0}^\infty(-1)^nt^n\,\mathrm dt=\sum_{n=1}^\infty\frac{(-1)^{n-1}}nx^n$$when $$x\in(-1,1)$$. Besides, it is clear that your series diverges when $$x=-1$$. Finally, the equality$$\log(2)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}n$$follows from Abel's theorem.