find-radius-of-S-x-n-1-x-n-n-2-and-calculate-its-sum-2-find-n-1-1-n-2-and-n-1-1-n-2-2-n-

Question Number 39517 by math khazana by abdo last updated on 07/Jul/18

f i n d r a d i u s o f S (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n^{2}}

a n d c a l c u l a t e i t s s u m

2) f i n d \sum_{n = 1}^{\infty} \frac{1}{n^{2}} a n d \sum_{n = 1}^{\infty} \frac{1}{n^{2} 2^{n}} .

Commented by abdo mathsup 649 cc last updated on 07/Jul/18

f o r x \neq 0 ∣ \frac{u_{n + 1} (x)}{u_{n} (x)} ∣ =∣ \frac{\frac{x^{n + 1}}{{(n + 1)}^{2}}}{\frac{x^{n}}{n^{2}}} ∣= \frac{n^{2}}{{(n + 1)}^{2}} ∣ x ∣\to∣ x ∣

s o i f ∣ x ∣< 1 t h e s e r i e c o n v e r g e s

i f x = - 1 Σ \frac{{(- 1)}^{n}}{n^{2}} c o n v e r g e s (a l t e r n a t e s e r i e)

i f x = 1 Σ \frac{1}{n^{2}} i s a r e i m a n s e r i e c o n v e r g e n t

a n d R = 1

w e h a v e \frac{d S}{d x} (x) = \sum_{n = 1}^{\infty} \frac{{n x}^{n - 1}}{n^{2}} = \frac{1}{x} \sum_{n = 1}^{\infty} \frac{x^{n}}{n}

= - \frac{l n (1 - x)}{x} (w e s u p p o s e t h a t x \neq 0) \Rightarrow

S (x) = - \int_{0}^{x} \frac{l n (1 - t)}{t} d t + c

c = S (0) = 0 \Rightarrow S (x) = - \int_{0}^{x} \frac{l n (1 - t)}{t} d t .

Commented by abdo mathsup 649 cc last updated on 07/Jul/18

2) \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = S (1) = - \int_{0}^{1} \frac{l n (1 - t)}{t} d t

a n d w e h a v e a l w a y s p r o v e d t h a t \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} .