Expand into power series $f(x)=\log(x+\sqrt{1+x^2})$ As in the topic, I am also supposed to find the radius of convergence. 
My solution: $$\log(x+\sqrt{1+x^2})=\log \left ( x(1+\sqrt{\frac{1}{x^2}+1})\right )=\log(x)+\log(1+\sqrt{\frac{1}{x^2}+1})$$Now I tried to use expansion for $\log(1+x)$ as $x\rightarrow0:\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$$
\log(1+(x-1))=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}(x-1)^k $$ $$\log(1+\sqrt{\frac{1}{x^2}+1})= \sqrt{\frac{1}{x^2}+1}-\frac{(\sqrt{\frac{1}{x^2}+1})^2}{2}+\frac{(\sqrt{\frac{1}{x^2}+1})^3}{3}-\cdots=$$$$=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}(\sqrt{\frac{1}{x^2}+1})^k$$Here I end up with nasty identity $$f(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}(x-1)^k+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}(\sqrt{\frac{1}{x^2}+1})^k$$and I don't know how to evaluate it into one serie. ($|x|<1$). Any hints? Thanks in advance
 A: Hint:
$$\dfrac{d}{dx}\ln\left(x+\sqrt{1+x^2}\right)=\dfrac{1}{\sqrt{1+x^2}}=(1+x^2)^{-\frac{1}{2}}.$$
Radius of convergence for the $\ln\left(x+\sqrt{1+x^2}\right)$ will be the same as for $(1+x^2)^{-\frac{1}{2}}.$
A: This is a lot easier to do if you instead consider the power series expansion of $\log(1-x)$.$$\log(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$$
We then manipulate this to yield $(\log(1-x))^2=\log(1-x)\log(1-x)$, so:$$(\log(1-x))^2=\left(\sum_{n=1}^\infty\frac{x^n}n\right)\left(\sum_{n=1}^\infty\frac{x^n}n\right)$$Have you ever toyed with multiplication of polynomials? Intuitively so, multiplying such power series is virtually identical. The coefficients of our product is the Cauchy product of the coefficients of our individual series.$$\begin{align*}(\log(1-x))^2&=\sum_{n=1}^\infty\left(\sum_{k=1}^n\frac1{k(n-k)}\right)x^n\\&=\dots\end{align*}$$
A: Your function, as a function on the complex numbers, is analytic at most points, and has branch points at
$$ z^2 + 1 = 0$$
due to the square root, and at
$$ z + \sqrt{z^2+1} = 0 $$
due to the logarithm.
The points satisfying the first equation are $\pm i$, and there are no solutions to the second equation.
(minor technical point: check that $\pm i$ truly are branch points, rather than simply being the only possibilities for branch points)
So the only singularities of your function are at $z=i$ and $z=-i$. The Taylor series for $f(z)$ around any point $a$, then, will have radius of convergence
$$ \min\{ |a-i|, |a+i| \} $$
