find-the-value-of-n-2-n-n-n-1-x-n-

Question Number 26567 by abdo imad last updated on 26/Dec/17

f i n d t h e v a l u e o f \sum_{n ⩾ 2} \frac{{(-)}^{n}}{n (n - 1)} x^{n}

Commented by prakash jain last updated on 27/Dec/17

a_{n} = \frac{{(- 1)}^{n + 1}}{n (n + 1)} x^{n + 1} n ⩾ 1

a_{n} = {(- 1)}^{n + 1} x^{n + 1} (\frac{1}{n} - \frac{1}{n + 1})

a_{n} = {(- 1)}^{n + 1} \frac{x^{n + 1}}{n} - {(- 1)}^{n + 1} \frac{x^{n + 1}}{n + 1}

S = x (x - \frac{x^{2}}{2} + \frac{x^{3}}{x} - + . .) +

(x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + - . .) - x

S = x \ln (1 + x) + \ln (1 + x) - x

- 1 < x ⩽ 1

Commented by abdo imad last updated on 28/Dec/17

l e t p u t S (x) = \sum_{n = 2}^{\propto} \frac{{(- 1)}^{n}}{n (n - 1)} x^{n} f o r / x / < 1

S (x) = \sum_{n ⩾ 2} {(- 1)}^{n} (\frac{1}{n - 1} - \frac{1}{n}) x^{n}

= - x \sum_{n ⩾ 2} \frac{{(- x)}^{n - 1}}{n - 1} - \sum_{n ⩾ 2} \frac{{(- x)}^{n}}{n}

= - x \sum_{n ⩾ 1} \frac{{(- x)}^{n}}{n} - \sum_{n ⩾ 1} \frac{{(- x)}^{n}}{n} - x

b u t {l n}^{,} (1 + u) = \frac{1}{1 + u} = Σ_{n ⩾ 0} {(- u)}^{n}

l n (1 + u) = Σ_{n ⩾ 0} {(- 1)}^{n} \frac{u^{n + 1}}{n + 1}

= \sum_{n ⩾ 1} {(- 1)}^{n - 1} \frac{u^{n}}{n} = - \sum_{n ⩾ 1} {(- 1)}^{n} \frac{u^{n}}{n}

\Rightarrow S (x) = x l n (1 + x) + l n (1 + x) - x

= (x + 1) l n x - x

Commented by abdo imad last updated on 28/Dec/17

y o u r a n s w e r i s t r u e p r a k a s h

Commented by abdo imad last updated on 28/Dec/17

S (x) = (x + 1) l n (x + 1) - x

Answered by prakash jain last updated on 28/Dec/17

f o r x = - 1

a_{n} = \frac{{(- 1)}^{2 n + 2}}{n (n + 1)} = \frac{1}{n (n + 1)} n ⩾ 1

= (\frac{1}{n} - \frac{1}{n + 1})

\underset{n = 1}{\sum^{\infty}} a_{n} = 1

Commented by prakash jain last updated on 27/Dec/17

\lim_{x \to - 1} (1 + x) \ln (1 + x) - x = 1