How do I calculate the sum of this series?∑n=1∞(−1)nn2n+1

∑n=1∞(−1)nn2n+1=12∑n=1∞qnn∣q=−12=12∑n=1∞∫0qtn−1dt∣q=−12=12∫0q11−tdt∣q=−12=12(−log⁡(1−q))∣q=−12=−12log⁡32

The Taylor series for log⁡(1+x) islog⁡(1+x)=∑k=1∞(−1)k−1xkkPluging in x=12, we getlog⁡(32)=∑k=1∞(−1)k−11k2kMultiplying by −12 yields−12log⁡(32)=∑k=1∞(−1)k1k2k+1

∑n=1∞(−1)nn2n+1=12∑n=1∞qnn∣q=−12=12∑n=1∞∫0qtn−1dt∣q=−12=12∫0q11−tdt∣q=−12=12(−log⁡(1−q))∣q=−12=−12log⁡32

The Taylor series for log⁡(1+x) islog⁡(1+x)=∑k=1∞(−1)k−1xkkPluging in x=12, we getlog⁡(32)=∑k=1∞(−1)k−11k2kMultiplying by −12 yields−12log⁡(32)=∑k=1∞(−1)k1k2k+1

Key to "How do I calculate the sum of this..."

Lorelei Stanton

Answered question

2022-03-25

How do I calculate the sum of this series?

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

Answer & Explanation

Mason Knight

Beginner2022-03-26Added 11 answers

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

= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{q^{n}}{n} ∣_{q = - \frac{1}{2}}

= \frac{1}{2} \sum_{n = 1}^{\infty} \int_{0}^{q} t^{n - 1} d t ∣_{q = - \frac{1}{2}}

= \frac{1}{2} \int_{0}^{q} \frac{1}{1 - t} d t ∣_{q = - \frac{1}{2}}

= \frac{1}{2} (- \log (1 - q)) ∣_{q = - \frac{1}{2}}

= - \frac{1}{2} \log \frac{3}{2}

okusen8m1a

Beginner2022-03-27Added 8 answers

The Taylor series for

\log (1 + x)

\log (1 + x) = \sum_{k = 1}^{\infty} {(- 1)}^{k - 1} \frac{x^{k}}{k}

Pluging in

x = \frac{1}{2}

, we get

\log (\frac{3}{2}) = \sum_{k = 1}^{\infty} {(- 1)}^{k - 1} \frac{1}{k 2^{k}}

Multiplying by

- \frac{1}{2}

yields

- \frac{1}{2} \log (\frac{3}{2}) = \sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{1}{k 2^{k + 1}}

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