Use-Abel-summation-to-evaluate-n-1-1-2n-1-2-n-1-2-ln-2-1-

Question Number 145827 by qaz last updated on 08/Jul/21

Use Abel summation to evaluate ::

\underset{n = 1}{\sum^{\infty}} \frac{1}{(2 n - 1) \cdot 2^{n}} = \frac{1}{\sqrt{2}} \ln (\sqrt{2} + 1)

Answered by Ar Brandon last updated on 08/Jul/21

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{(2 n - 1) 2^{n}} = \underset{n = 1}{\sum^{\infty}} \frac{1}{(2 n - 1) {(\sqrt{2})}^{2 n - 1} \cdot \sqrt{2}}

f (a) = \underset{n = 1}{\sum^{\infty}} \frac{a^{2 n - 1}}{2 n - 1} \Rightarrow f^{'} (a) = \underset{n = 1}{\sum^{\infty}} a^{2 n - 2} = \frac{1}{1 - a^{2}}, ∣ a ∣< 1

f (a) = \frac{1}{2} \ln ∣ \frac{1 + a}{1 - a} ∣ + C, f (0) = 0 \Rightarrow C = 0

f (a) = \frac{1}{2} \ln (\frac{1 + a}{1 - a})

S = \frac{1}{\sqrt{2}} f (\frac{1}{\sqrt{2}}) = \frac{1}{2 \sqrt{2}} \ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) = \frac{1}{2 \sqrt{2}} \ln {(\sqrt{2} + 1)}^{2}

= \frac{1}{\sqrt{2}} \ln (\sqrt{2} + 1)

Answered by mathmax by abdo last updated on 09/Jul/21

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

= \frac{1}{\sqrt{2}} f (\frac{1}{\sqrt{2}}) with f (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{2 n - 1} \Rightarrow f^{^{'}} (x) = \sum_{n = 1}^{\infty} x^{2 n - 2} and ∣ x ∣< 1

= \sum_{n = 0}^{\infty} x^{2 n} = \frac{1}{1 - x^{2}} \Rightarrow f (x) = \int \frac{dx}{1 - x^{2}} + k

= \frac{1}{2} \int (\frac{1}{1 - x} + \frac{1}{1 + x}) dx + K = \frac{1}{2} \log ∣ \frac{1 + x}{1 - x} ∣ + K

f (0) = 0 = k \Rightarrow f (x) = \frac{1}{2} \log ∣ \frac{1 + x}{1 - x} ∣ \Rightarrow S = \frac{1}{2 \sqrt{2}} \log ∣ \frac{1 + \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} ∣

= \frac{1}{2 \sqrt{2}} \log (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})