1-S-n-1-1-n-1-n-2-n-2-A-1-n-1-n-2-2-n-

Question Number 146736 by mnjuly1970 last updated on 15/Jul/21

1 : S := \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n . 2^{n}} = ?

2 : A := Σ \frac{{(- 1)}^{n - 1}}{n^{2} . 2^{n}} = ?

Answered by Olaf_Thorendsen last updated on 15/Jul/21

\frac{1}{1 + x} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{n}

\ln (1 + x) = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1}

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

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

\int \frac{\ln (1 + x)}{x} d x = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n - 1} \frac{x^{n}}{n^{2}} = - {Li}_{2} (1 + x) (2)

(1) : S = \ln (1 + \frac{1}{2}) = \ln \frac{3}{2}

(2) : A = - {Li}_{2} (1 + \frac{1}{2}) = - {Li}_{2} (\frac{3}{2})

Commented by mnjuly1970 last updated on 15/Jul/21

t h a n k s a l o t m a s t e r \dots .