Solving Maclaurin series

Question

I have to tackle the following question. My thoughts so far are below it.

(A) By successive differentiation find the first four non-zero terms in the Maclaurin series for $$F(x)=(x+1)\ln(1+x)-x$$ (B) Deduce that, for $n\geq2$, the coefficient of $X^n$ in this series is $(-1)^n\cdot(1/n(n-1))$

(C) By applying the ratio test, find the radius of convergence for this Maclaurin series

I got $$F(x)=(1/2)x^2-(1/6)x^3+(1/12)x^4-(1/20)x^5$$ for part A.
I can easily find the general coefficient in part B: $$(-1)^n \cdot (n－2)!/n! = (-1)^n ·1/n(n-1),$$ but my problem is, what is the right format to deduce?
For part C, I am very confused because after applying the ratio test my result is $1,$ how can I find the radius of convergence? Help! Thanks!

Answer 1

Hint:

$$F'(x)=\ln(x+1)$$ and then,

$$F^{(n)}(x)=(-1)^n\frac{(n-2)!}{(x+1)^{n-1}},$$

and

$$F^{(n)}(0)=(-1)^n(n-2)!.$$

Hence the $n^{th}$ coefficient is

$$(-1)^n\frac{(n-2)!}{n!}.$$

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Applying the ratio test,

$$\lim_{n\to\infty}\frac{n(n-1)}{(n+1)n}=1$$ which is the radius of convergence.

Note that for $x=\pm1$, the terms decrease like $n^{-2}$ so that the series also converges.

Answer 2

Just for fun, another method:

We know the development

$$\ln(1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}.$$

Then,

$$(1+x)\ln(1+x)=\sum_{n=1}^\infty\left((-1)^{n+1}\frac{x^n}{n}+(-1)^{n-1}\frac{x^{n+1}}{n}\right)=x+\sum_{n=2}^\infty(-1)^{n}\left(-\frac1{n}+\frac{1}{n-1}\right)x^n.$$