Taylor series $\ln(2+x)$ centered at $x=2$

Question

Taylor series $\ln(2+x)$ centered at $x=2$.

Is the correct result $$y=\ln \left(4\right)+\sum _{n=1}^{∞}\frac{\left(-1\right)^n}{4^{\left(2^{\Large n}\right)}}\cdot \frac{\left(x-2\right)^n}{n!}\ ?$$

Thanks.

Answer 1

You probably started right, using $x+2=4+x-2=4\left(1+\frac{x-2}{4}\right)$.

So we want $\ln 4+\ln\left(1+\frac{x-2}{4}\right)$.

Now use the series for $\ln(1+t)$, and replace $t$ everywhere by $\frac{x-2}{4}$.