Expand $f(x)=\log(x+ \sqrt{1+x^2})$ into power series and determine convergence intervals

Question

We have $f(x)=\log(x+ \sqrt{1+x^2})$ and we need to expand it into power series, which I suppose is easy because $f'(x)=(1+x^2)^{-1/2}= \sum_{k=0} {-\frac{1}{2} \choose k}x^{2k}$. It follows that $f(x)=c+\sum_{k=0} {-\frac{1}{2} \choose k}\frac{x^{2k+1}}{2k+1}$, where $c$ is some constant.

Using ratio test it is easy to determine radius of convergence, so when $x \in (-1,1)$, then series converges. Now we have to check end points and that's where I don't know how to proceed.

Answer 1

As you noticed, we have: $$ f'(x) = \frac{1}{\sqrt{1+x^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}(-1)^n x^{2n} $$ and since $f(0)=0$, $$ f(x)=\log\left(x+\sqrt{1+x^2}\right) = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\cdot\frac{(-1)^n}{2n+1}\cdot x^{2n+1} $$ where convergence at $x=\pm 1$ is granted, since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$ for large $n$.

Answer 2

Expand $(1+x^2)^{-1/2}=1-\frac{x^2}{2}+\frac{1.3x^4}{2^2.2!}-\frac{1.3.5x^6}{2^3.3!}+\frac{1.3.5.7x^8}{2^4.4!}-...$ using bionomial theorem

$(1+x^2)^{-1/2}= \frac{1+\frac{2x}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}$ $=1-\frac{x^2}{2}+\frac{1.3x^4}{2^2.2!}-\frac{1.3.5x^6}{2^3.3!}+\frac{1.3.5.7x^8}{2^4.4!}-...$

Now integration term by term will do the trick. We will get

$log(1+\sqrt{1+x^2}) =x-\frac{x^3}{2.3}+\frac{1.3x^5}{2^2.2!.5}-\frac{1.3.5x^7}{2^3.3!.7}+\frac{1.3.5.7x^9}{2^4.4!.9}-...$