Explicit-f-a-n-1-1-n-n-an-1-

Question Number 68149 by ~ À ® @ 237 ~ last updated on 06/Sep/19

E x p l i c i t f (a) = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n (a n + 1)}

Commented by turbo msup by abdo last updated on 06/Sep/19

i f a = 0 f (0) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n}

= - l n (2)

i f a \neq 0 w e h a v e \frac{f (a)}{a} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n (a n + 1)}

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

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

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

l e t s (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n + 1} x^{a n + 1} w i t h ∣ x ∣< 1

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

s^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{a n}

= \sum_{n = 1}^{\infty} {(- x^{a})}^{n} = \frac{1}{1 + x^{a}} \Rightarrow

s (x) = \int_{0}^{x} \frac{d t}{1 + t^{a}} + c

s (0) = 0 = c \Rightarrow s (x) = \int_{0}^{x} \frac{d t}{1 + t^{a}} \Rightarrow

s (1) = \int_{0}^{1} \frac{d t}{1 + t^{a}} \Rightarrow

\frac{f (a)}{a} = - \frac{1}{a} l n (2) - \int_{0}^{1} \frac{d t}{1 + t^{a}} \Rightarrow

f (a) = - l n (2) - a \int_{0}^{1} \frac{d t}{1 + t^{a}}

b e c o n t i n u e d \dots .

Commented by mathmax by abdo last updated on 06/Sep/19

e r r o r f r o m l i n e 10 s^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- x^{a})}^{n} = - x^{a} \sum_{n = 1}^{\infty} {(- x^{a})}^{n - 1}

= - x^{a} \sum_{n = 0}^{\infty} {(- x^{a})}^{n} = \frac{- x^{a}}{1 + x^{a}} \Rightarrow s (x) = - \int_{0}^{x} \frac{t^{a}}{1 + t^{a}} d t + c

s (0) = 0 = c \Rightarrow s (x) = - \int_{0}^{x} \frac{t^{a}}{1 + t^{a}} d t \Rightarrow

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n + 1} = s (1) = - \int_{0}^{1} \frac{t^{a}}{1 + t^{a}} d t = - \int_{0}^{1} \frac{1 + t^{a} - 1}{1 + t^{a}} d t

= - 1 + \int_{0}^{1} \frac{d t}{1 + t^{a}} b e c o n t i n u e d \dots