Get the answer to "Prove that:∑n=1∞(−1)n+1n(n+1)=2ln⁡2−1"

trasbocohf1

Answered question

2022-02-24

Prove that:

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n (n + 1)} = 2 \ln 2 - 1

husudiwareh

Beginner2022-02-25Added 7 answers

\sum_{k = 1}^{n} \frac{{(- 1)}^{k + 1}}{k (k + 1)} = \sum_{k = 1}^{n} {(- 1)}^{k + 1} (\frac{1}{k} - \frac{1}{k + 1}) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k + 1}}{k} + \sum_{k = 1}^{n} \frac{{(- 1)}^{k + 2}}{k + 1}

= 2 \sum_{k = 1}^{n} \frac{{(- 1)}^{k + 1}}{k} - 1 - \frac{{(- 1)}^{n + 1}}{n + 1}

Hence,

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n (n + 1)} = 2 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n} - 1 = 2 \ln 2 - 1

since

\sum_{n = 1}^{n} \frac{{(- 1)}^{n + 1}}{n} = \ln 2

Proof. If

| x | < 1

, then

\frac{1}{1 + x} = \sum_{k = 0}^{n} {(- 1)}^{k} x^{k} + \frac{{(- 1)}^{n + 1} x^{n + 1}}{1 + x}

and hence

\log (1 + x) = \int_{0}^{x} \frac{d t}{1 + t} = \sum_{k = 0}^{n} {(- 1)}^{k} \int_{0}^{t} t^{k} d t + \int_{0}^{x} \frac{{(- 1)}^{n + 1} t^{n + 1} d t}{1 + t}

= \sum_{k = 0}^{n} \frac{{(- 1)}^{k} x^{n + 1}}{n + 1} + R_{n} (x)

Clearly, for

x \in [0, 1]

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