Explicit f(a)=Σ_(n=1) ^∞ (((−1)^n )/(n(an+1)))


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
$if a=0 f(0)=Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) if a≠0 we have ((f(a))/a)=Σ_(n=1) ^∞ (((−1)^n )/(an(an+1))) =Σ_(n=1) ^∞ (−1)^n {(1/(an))−(1/(an+1))} =(1/a)Σ_(n=1) ^∞ (((−1)^n )/n)−Σ_(n=1) ^∞ (((−1)^n )/(an+1)) =−(1/a)ln(2)−Σ_(n=1) ^∞ (((−1)^n )/(an+1)) let s(x) =Σ_(n=1) ^∞ (((−1)^n )/(an+1))x^(an+1) with∣x∣<1 s(1)=Σ_(n=1) ^∞ (((−1)^n )/(an+1)) s^′ (x) =Σ_(n=1) ^∞ (−1)^n x^(an) =Σ_(n=1) ^∞ (−x^a )^n =(1/(1+x^a )) ⇒ s(x) =∫_0 ^x (dt/(1+t^a )) +c s(0)=0=c ⇒s(x)=∫_0 ^x (dt/(1+t^a )) ⇒ s(1) =∫_0 ^1 (dt/(1+t^a )) ⇒ ((f(a))/a) =−(1/a)ln(2)−∫_0 ^1 (dt/(1+t^a )) ⇒ f(a) =−ln(2)−a∫_0 ^1 (dt/(1+t^a )) be continued....$
$i f a = 0 f (0) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n}$ $= - l n (2)$ $i f a \neq 0 w e h a v e \frac{f (a)}{a} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n (a n + 1)}$ $= \sum_{n = 1}^{\infty} {(- 1)}^{n} {\frac{1}{a n} - \frac{1}{a n + 1}}$ $= \frac{1}{a} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n + 1}$ $= - \frac{1}{a} l n (2) - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n + 1}$ $l e t s (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n + 1} x^{a n + 1} w i t h ∣ x ∣< 1$ $s (1) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n + 1}$ $s^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{a n}$ $= \sum_{n = 1}^{\infty} {(- x^{a})}^{n} = \frac{1}{1 + x^{a}} \Rightarrow$ $s (x) = \int_{0}^{x} \frac{d t}{1 + t^{a}} + c$ $s (0) = 0 = c \Rightarrow s (x) = \int_{0}^{x} \frac{d t}{1 + t^{a}} \Rightarrow$ $s (1) = \int_{0}^{1} \frac{d t}{1 + t^{a}} \Rightarrow$ $\frac{f (a)}{a} = - \frac{1}{a} l n (2) - \int_{0}^{1} \frac{d t}{1 + t^{a}} \Rightarrow$ $f (a) = - l n (2) - a \int_{0}^{1} \frac{d t}{1 + t^{a}}$ $b e c o n t i n u e d . . . .$
Commented by mathmax by abdo last updated on 06/Sep/19

$e r r o r f r o m l i n e 10 s^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- x^{a})}^{n} = - x^{a} \sum_{n = 1}^{\infty} {(- x^{a})}^{n - 1}$ $= - x^{a} \sum_{n = 0}^{\infty} {(- x^{a})}^{n} = \frac{- x^{a}}{1 + x^{a}} \Rightarrow s (x) = - \int_{0}^{x} \frac{t^{a}}{1 + t^{a}} d t + c$ $s (0) = 0 = c \Rightarrow s (x) = - \int_{0}^{x} \frac{t^{a}}{1 + t^{a}} d t \Rightarrow$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{a n + 1} = s (1) = - \int_{0}^{1} \frac{t^{a}}{1 + t^{a}} d t = - \int_{0}^{1} \frac{1 + t^{a} - 1}{1 + t^{a}} d t$ $= - 1 + \int_{0}^{1} \frac{d t}{1 + t^{a}} b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum