find-n-0-n-1-x-3n-2-calculate-n-0-n-1-8-n-

Question Number 33709 by math khazana by abdo last updated on 22/Apr/18

f i n d \sum_{n = 0}^{\infty} (n + 1) x^{3 n}

2) c a l c u l a t e \sum_{n = 0}^{\infty} \frac{n + 1}{8^{n}} .

Commented by prof Abdo imad last updated on 27/Apr/18

l e t p u t f o r ∣ x ∣< 1 w (x) = \sum_{n = 0}^{\infty} (n + 1) x^{3 n} w e h a v e

x^{3} w (x) = \sum_{n = 0}^{\infty} (n + 1) {(x^{3})}^{n + 1} = φ (x^{3}) w i t h

φ (t) = \sum_{n = 0}^{\infty} (n + 1) t^{n + 1} b u t w e h a v e

\sum_{n = 0}^{\infty} t^{n} = \frac{1}{1 - t} \Rightarrow \sum_{n = 1}^{\infty} n t^{n - 1} = \frac{1}{{(1 - t)}^{2}} \Rightarrow

\sum_{n = 1}^{\infty} n t^{n} = \frac{t}{{(1 - t)}^{2}} b u t φ (t) = \sum_{n = 1}^{\infty} n t^{n} \Rightarrow

φ (t) = \frac{t}{{(1 - t)}^{2}} w i t h ∣ t ∣< 1 \Rightarrow x^{3} w (x) = \frac{x^{3}}{{(1 - x^{3})}^{2}}

\Rightarrow w (x) = \frac{1}{{(1 - x^{3})}^{2}}

2) w e h a v e \sum_{n = 0}^{\infty} \frac{n + 1}{8^{n}} = \sum_{n = 0}^{\infty} (n + 1) {(\frac{1}{2})}^{3 n}

= w (\frac{1}{2}) = \frac{1}{{(1 - \frac{1}{8})}^{2}} = \frac{1}{{(\frac{7}{8})}^{2}} = \frac{64}{49} .