let-give-S-x-n-1-x-n-n-and-W-x-n-1-1-n-x-n-n-2-calculate-S-x-W-x-in-that-we-know-x-lt-1-

Question Number 27098 by abdo imad last updated on 02/Jan/18

l e t g i v e S (x) = \sum_{n = 1}^{\propto} \frac{x^{n}}{n} a n d W (x) = \sum_{n = 1}^{\propto} \frac{{(- 1)}^{n} x^{n}}{n^{2}}

c a l c u l a t e S (x) . W (x) . i n t h a t w e k n o w / x / < 1 .

Commented by prakash jain last updated on 02/Jan/18

\ln (1 - x) = - x - \frac{x^{2}}{2} - \dots (A)

= - \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n}

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

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

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

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

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

W (x) = \frac{(x + 1) \ln (1 + x) - x}{x}

Given ∣ x ∣< 1 both S (x) and W (x)

converge .

Commented by abdo imad last updated on 04/Jan/18

l e t c a l c u l a t e S (x) W (x) i n f o r m o f s e r i e s w e p u t a_{n} = \frac{1}{n}

a n d b_{n} = \frac{{(- 1)}^{n}}{n^{2}}

S (x) W (x) = \sum_{n = 1}^{\propto} c_{n} x^{n} w i t h c_{n} = \sum_{i + j = n} a_{i} b_{j}

c_{n} = \sum_{i = 1}^{n - 1} a_{i} b_{n - i}

S (x) W (x) = \sum_{n = 1}^{\propto} (\sum_{i = 1}^{n - 1} \frac{1}{i} \frac{{(- 1)}^{n - i}}{{(n - i)}^{2}}) x^{n} .