Find a power series representation for f(x)=ln(x+1) centered at 1

Question

Is my answer correct?

$$f(x)=ln(x+1)$$ $$\int\frac{1}{x+1}dx$$ $$\int\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}{(x-1)^n}dx$$ $$\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}\frac{(x-1)^{n+1}}{n+1}$$ $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n2^n}{(x-1)}^n$$

Answer 1

Yes, your answer is correct. Just to be clear for future readers, you've correctly noticed that $$\frac{1}{x+1}=\frac{1}{2+(x-1)}=\frac{1}{2}\cdot\frac{1}{1-\left(-\frac{x-1}{2}\right)}.$$ At this point, you've used the formula for a geometric series: $\frac{1}{1-y}=\sum_{n=0}^\infty y^n$ and correctly simplified.