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Is my answer correct?

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

Zoey
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    Looks good. You should specify domain: $|\frac{x-1}{2}|\lt 1$. – herb steinberg May 02 '19 at 00:08
  • Oops I also forgot to add +C to the equation. To solve for C would I plug in x=1 since that's the center? – Zoey May 02 '19 at 00:38
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    Possible duplicate of Represent f(x)=1/(1+x) as a power series around x=1 – Clayton May 02 '19 at 01:58
  • Why did you make essentially the same question twice? To verify you have the correct solution to someone's hint, it is better etiquette to post a comment under their answer and ask the author of the hint (or under each answer) as opposed to creating an entire question to ask if you understood someone's hint. – Clayton May 02 '19 at 02:00
  • @Clayton Sorry I just signed up for this today. My questions were both separate questions on my assignment, so I addressed them separately. In the first, I was simply asking for how to solve for the power series with 1/(1+x). In the second, I was asking about f(x)=ln(x+1), which to me seemed like a very different question (that happened to use the power series from my other question). They are similar, but not the same. Nevertheless, I will combine the two as was advised. Thanks. – Zoey May 02 '19 at 03:16

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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.

Clayton
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