<s>[INST] series convergence proof. Use Theorem 8.8 to show that if $0<x\leq 1$ then ln$(1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}x^k$. 

Theorem 8.8 states that for each natural number $n$ and each number $x>-1$ there is a number $c$ strictly between 0 and $x$ such that ln$(x+1)=x-\frac{x^2}{2}+..+\frac{(-1)^{n+1}x^n}{n}+\frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}$

Proof: 
Assume that $0<x\leq 1$ and it follows from theorem 8.8 that ln$(x+1)=\sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}+\frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}\implies \lim_{n\to\infty} ln(1+x)=\lim_{n\to\infty} \sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}+\lim_{n\to\infty} \frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}$ for some $c$ between $0$ and $x$. We see that  $\lim_{n\to\infty} ln(1+x)=ln(1+x)$ and $\lim_{n\to\infty} \frac{(-1)^n}{(n+1)(1+c)^{n+1}}x^{n+1}=0$. Also $\lim_{n\to\infty} \sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}{k} $ by definition. Thus $ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}{k}$. Am I on the right track here or no? Thanks
 [/INST] I think you are on the right track. I reorganize your proof here, and add one or two details. For $x \in (0, 1]$, let $S_n(x) = \displaystyle \sum_{k=1}^n \dfrac{(-1)^{k+1}x^k}{k}$, then you need to prove: $\displaystyle \lim_{n \to \infty} \left(S_n(x) - ln(1+x)\right) = 0$, and using theorem 8.8, we have: $S_n(x) - ln(1+x) = \dfrac{1}{n+1}\cdot \left(\dfrac{-x}{1+c}\right)^{n+1}$. Next since $-1 < \dfrac{-x}{1+c} < 1$. So:  $0 < \big|\dfrac{1}{n+1}\cdot \left(\dfrac{-x}{1+c}\right)^{n+1}\big| < \dfrac{1}{n+1}$, so by squeeze theorem, $\displaystyle \lim_{n \to \infty} \left(S_n(x) - ln(1+x)\right) = 0$,and the statement is proved. 
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