find-the-ranges-of-value-of-x-for-which-series-convergent-or-divergent-n-1-n-1-n-3-x-n-

Question Number 203329 by Calculusboy last updated on 16/Jan/24

f i n d t h e r a n g e s o f v a l u e o f x f o r w h i c h

s e r i e s c o n v e r g e n t o r d i v e r g e n t

\underset{n = 1}{\sum^{\infty}} \frac{(n + 1)}{n^{3}} x^{n}

Answered by Mathspace last updated on 17/Jan/24

u_{n} (x) = \frac{n + 1}{n^{3}} x^{n}

∣ \frac{u_{n + 1} (x)}{u_{n} (x)} ∣=∣ \frac{\frac{n + 2}{{(n + 1)}^{3}} x^{n + 1}}{\frac{n + 1}{n^{3}} x^{n}} ∣

= \frac{n + 2}{{(n + 1)}^{3}} \times \frac{n^{3}}{n + 1} ∣ x ∣

= {(\frac{n}{n + 1})}^{3} . \frac{n + 2}{n + 1} ∣ x ∣\to∣ x ∣ (n \to + \infty)

s i ∣ x ∣< 1 t h e s e r i e c v a n d r = 1

s i ∣ x ∣> 1 t h e s e r i e d i v e r g e s

i f x = 1 u_{n} (x) = \frac{n + 1}{n^{3}} \sim \frac{1}{n^{2}}

Σ \frac{1}{n^{2}} c v \Rightarrow Σ u_{n} (x) c v

i f = x = - 1 u_{n} (x) = \frac{(n + 1)}{n^{3}} {(- 1)}^{n}

Σ u_{n} (x) i s a a l t e r n a t i n g s e r i e

\Rightarrow Σ u_{n} (x) c v .

Commented by Calculusboy last updated on 17/Jan/24

n i c e s i r