Taylor series for $-\log(1-t)$

Question

Have you ever come across such a proof? I have no idea how to solve that problem:
"Prove that for $t \in (-1, 1)$:
$-\log(1-t) = \lim_{n\to\infty}\big(\frac{t}{1} + \frac{t^{2}}{1} + \frac{t^{3}}{1} + ... + \frac{t^{n}}{n}\big)$".
There is a tip to use Cauchy's rest which is:
$R_{n}(t) = \frac{f^{(n-1)}(s)}{(n-1)!}(t-s)^{n-1}(t-a)$

Answer 1

Here $f (t) =\ln (1-t) $, therefore $f'(t) =-\frac{1}{1-t}, f''(t) = -\frac{1}{(1-t)^2},f'''(t) = -\frac{1}{(1-t)^3 }$, etc.

Maclaurin's series expansion is $$f (t) =f (0)+tf'(0) +\frac {t^2}{2!}f''(0) +\cdots $$ Can you take it from here?

Answer 2

If you have covered derivation of series $$f(t):=t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\cdots+\dfrac{t^n}{n}+\cdots$$ $$\Rightarrow f'(t)=1+t+t^2+\cdots+t^{n-1}+\cdots=\frac{1}{1-t},\quad \left|t\right|<1$$ $$\Rightarrow f(t)=\int \frac{dt}{1-t}=-\log (1-t)+C.$$ For $t=0$ an taking into account that $f(0)=0$ we get $0=(-\log 1)+C,$ so $C=0.$ Then, $$-\log (1-t)=\underbrace{t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\cdots+\dfrac{t^n}{n}}_{S_n(t)}+\underbrace{\cdots}_{R_n(t)}$$ $$\Rightarrow\lim_{n\to +\infty}\left(t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\cdots+\dfrac{t^n}{n}\right)=-\log (1-t),\quad |t|<1.$$