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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:
$R_{n}(t) = \frac{f^{(n-1)}(s)}{(n-1)!}(t-s)^{n-1}(t-a)$

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    why not just integrate ${1\over 1-x}$? – 2017-01-09
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    Then I would get $\log(1-x)$ and what next? – 2017-01-09
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    I'm really sorry. Just a mistake while typing. It's fixed. – 2017-01-09
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    you would get $-\log(1-x)$ and then integrate the Taylor series of ${1\over 1-x}$ you get your desired series. – 2017-01-09
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    That's a smart way of solving it. Thanks! – 2017-01-09

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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, $$-\log (1-t)=\underbrace{t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\cdots+\dfrac{t^n}{n}}_{S_n(t)}+\underbrace{\cdots}_{R_n(t)}$$ $$\Rightarrow\lim_{n\to +\infty}\left(t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\cdots+\dfrac{t^n}{n}\right)=-\log (1-t),\quad |t|<1.$$

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    That's a nice solution. Thanks! – 2017-01-09
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    What is $S_{n}(t)$? – 2017-01-09
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    And why $f(t) = t + \frac{t^{2}}{2} + ... + \frac{t^{n}}{n} + ...$ that is not Taylor's series expansion for $-\log(1-t)$ – 2017-01-09
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    @Hendra $S_n(t)$ is the nth partial sum, and $R_n(t)$ the remainder. As $f(t)=S_n(t)+R_n(t)$ and the series $t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\cdots+\dfrac{t^n}{n}+\cdots$ is convergent for $|t|<1$ then, $R_n(t)\to 0$ as $n\to +\infty$ – 2017-01-09
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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?

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    I thought about it but how can I get $\lim_{n\to\infty}$ from Maclaurin's series expansion? – 2017-01-09
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    Finally $$f (t) = -t-\frac{t^2}{2!}-\frac {t^3}{3!}- \cdots $$ – 2017-01-09
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    But we were to get: $\frac{t^{n}}{n}}$ not $\frac{t^{n}}{n!}}$? Maybe it is just a mistake in my book. – 2017-01-09