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enthral4kvri

Answered question

2022-06-01

Proof verification

\sum_{n = 1}^{\infty} (- 1)^{n} \frac{1}{n}

converges

Answer & Explanation

Selena Pratt

Beginner2022-06-02Added 5 answers

I propose you this proof that does not use the criteria :

\begin{aligned} \sum_{n = 1}^{N} \frac{(- 1)^{n + 1}}{n} & = - \sum_{n = 1}^{N} \frac{(- 1)^{n}}{n} \\ = - \sum_{n = 1}^{N} (- 1)^{n} \int_{0}^{1} x^{n - 1} d x \\ = \int_{0}^{1} \sum_{n = 1}^{N} (- x)^{n - 1} d x \\ = \int_{0}^{1} \frac{1 - (- x)^{N}}{1 - (- x)} d x \\ = \int_{0}^{1} \frac{1}{1 + x} d x - \int_{0}^{1} \frac{(- x)^{N}}{1 + x} d x \\ = \ln (2) - \int_{0}^{1} \frac{(- x)^{N}}{1 + x} d x \end{aligned}

Then we can show that the second integral converges towards 0 when N grows to infinity :

| \int_{0}^{1} \frac{(- x)^{N}}{1 + x} d x | \leq \int_{0}^{1} x^{N} d x = \frac{1}{N + 1} \to 0

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