Two Equivalent Series for $ \ln(x^2+4)$

Question

I was asked to find the corresponding series for the function $\ln(x^2+4)$

The obvious solution to me was to use the well known fact $$\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}$$ And substituting $x^2+3$ for $x$ $$\ln(1+(x^2+3))=\sum_{n=1}^\infty (-1)^{n-1}\frac{(x^2+3)^n}{n}$$ Using binomial theorem on the $(x^2+3)^n$ on the inside gives us the nested summation $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\sum_{m=0}^n {n\choose{m}}x^{2m}3^{n-m}=S_1$$

However, the answer key gives the series as $$S_2=\ln 4+\sum_{n=1}^\infty (-1)^n \frac{x^{2n+2}}{2^{2n+2}(n+1)}$$

Question: Is $S_1=S_2$? If so, how do we prove this? If not, where is the error in this reasoning?

Thanks

Answer 1

The series $$\ln(1+t)=\sum_{n=1}^\infty(-1)^{n-1}\frac{t^n}n$$ is only valid for $|t|<1$. You apply it for $t=x^2+3$. I don't think $|x^2+3|<1$.

Answer 2

HINT: $$ 4+x^2=4\left(1+\frac{x^2}{4}\right) $$ and $\ln ab=\ln a+\ln b$.