calculate-k-1-1-k-1-99k-1-

Question Number 65386 by mathmax by abdo last updated on 29/Jul/19

c a l c u l a t e \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{99 k - 1}

Commented by mathmax by abdo last updated on 31/Jul/19

l e t I = \int_{1}^{+ \infty} \frac{d x}{1 + x^{99}} c h a 7 g e m e n t x = \frac{1}{t} g i v e

I = - \int_{0}^{1} \frac{1}{1 + \frac{1}{t^{99}}} (- \frac{d t}{t^{2}}) = \int_{0}^{1} \frac{t^{99}}{t^{2} (1 + t^{99})} d t = \int_{0}^{1} \frac{t^{97}}{1 + t^{99}} d t

= \int_{0}^{1} t^{97} \sum_{k = 0}^{\infty} {(- 1)}^{k} t^{99 k} d t = \sum_{k = 0}^{\infty} {(- 1)}^{k} \int_{0}^{1} t^{99 k + 97} d t

= \sum_{k = 0}^{\infty} {(- 1)}^{k} {[\frac{1}{99 k + 98} t^{99 k + 98}]}_{0}^{1} = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{99 k + 98}

=_{k = p - 1} \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{99 (p - 1) + 98} = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{99 p - 1} \Rightarrow

\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{99 k - 1} = \int_{1}^{+ \infty} \frac{d x}{1 + x^{99}}

\int_{1}^{+ \infty} \frac{d x}{1 + x^{99}} = \int_{1}^{0} (\dots .) d x + \int_{0}^{\infty} \frac{d x}{1 + x^{99}} = \int_{0}^{\infty} \frac{d x}{1 + x^{99}} - \int_{0}^{1} \frac{d x}{1 + x^{99}}

c h a n g e m e n t x = t^{\frac{1}{99}} g i v e \int_{0}^{\infty} \frac{d x}{1 + x^{99}} = \int_{0}^{\infty} \frac{1}{1 + t} \frac{1}{99} t^{\frac{1}{99} - 1} d t

= \frac{1}{99} \int_{0}^{\infty} \frac{t^{\frac{1}{99} - 1}}{1 + t} d t = \frac{1}{99} \frac{π}{s i n (\frac{π}{99})} = \frac{π}{99 s i n (\frac{π}{99})}

\int_{0}^{1} \frac{d x}{1 + x^{99}} = \int_{0}^{1} \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{99 n} d x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{99 n} d x

= \sum_{n = 0}^{\infty} {(- 1)}^{n} {[\frac{1}{99 n + 1} x^{99 n + 1}]}_{0}^{1} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{99 n + 1} = S

= 1 - \frac{1}{100} + \frac{1}{199} - \frac{1}{3.99 + 1} + \dots . \Rightarrow

\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{99 k - 1} = \frac{π}{99 s i n (\frac{π}{99})} - S