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Given $\frac{1}{1+x}=\sum^{\infty}_{k=0}(1-x)x^{2k}$ for $x\in[0,1)$ I shall apply the 'monotone convergence theorem' on $[0,1)$ for calculating $\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}$.

In Wikipedia it's said that $\sum^{\infty}_{k=0}\frac{1}{n^k}\binom{n}{k}=(1+\frac{1}{n})^n$ and I don't know how they apply this theorem on it :/ Maybe someone can explain me it with the example from Wikipedia :)

I know that $\int_0^1\frac{1}{1+x}dx=[ln(x)]_0^1$, but don't know how to put this in relation with the second sum ? Thanks in advance.

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Note that we can write

$$\begin{align} \int_0^1 \sum_{k=0}^\infty (1-x)x^{2k}\,dx&=\int_0^1\lim_{K\to \infty} \left(\frac{1-x^{2K+2}}{1+x}\right)\,dx\\\\ &\overbrace{=}^{\text{MCT}}\lim_{K\to \infty} \int_0^1\left(\frac{1-x^{2K+2}}{1+x}\right)\,dx\\\\ &=\sum_{k=0}^\infty \int_0^1 (1-x)x^{2k}\,dx \\\\ &=\sum_{k=0}^\infty \left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\\\\ &=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} \end{align}$$

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    I have some questions: 1) shouldn't it be $\frac{1-x^{2K+2}}{1+x}$ instead of $\frac{1-x^{2K+1}}{1+x}$ 2) The MCT says that we can change limit and integral right ? 3) The third equality why did you rewrite it in this way ? Couldn't we do this in the first equailty ? – 2017-01-18
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    By the way - thank you very much for your time and answer ! – 2017-01-18
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    You're welcome. It was my pleasure. On Question 1, you're correct; I've edited the typo. On Question 2, the MCT permits the interchange. On Question 3, we need to justify interchanging the series and integral. The second step justifies this and the third moves the sum outside the integral. – 2017-01-18