calculate-i-j-N-2-i-j-3-i-j-

Question Number 46607 by maxmathsup by imad last updated on 29/Oct/18

c a l c u l a t e \sum_{(i, j) \in N^{2}} \frac{i + j}{3^{i + j}}

Commented by maxmathsup by imad last updated on 30/Oct/18

w e h a v e S_{n} = \sum_{i = 0}^{\infty} (\sum_{j = 0}^{\infty} \frac{i + j}{3^{i + j}}) = \sum_{i = 0}^{\infty} S_{i}

S_{i} = \frac{i}{3^{i}} \sum_{j = 0}^{\infty} \frac{1}{3^{j}} + \frac{1}{3^{i}} \sum_{j = 0}^{\infty} \frac{j}{3^{j}} b u t \sum_{j = 0}^{\infty} \frac{1}{3^{j}} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} a l s o

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

w e h a v e \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} f o r ∣ x ∣< 1 \Rightarrow \sum_{n = 1}^{\infty} {n x}^{n - 1} = \frac{x}{{(1 - x)}^{2}} \Rightarrow

\sum_{n = 1}^{\infty} {n x}^{n} = \frac{x^{2}}{{(1 - x)}^{2}} \Rightarrow w (\frac{1}{3}) = \frac{1}{9 {(\frac{2}{3})}^{2}} = \frac{1}{4} \Rightarrow S_{i} = \frac{3}{2} \frac{i}{3^{i}} + \frac{1}{4 . 3^{i}} \Rightarrow

\sum_{i = 0}^{\infty} S_{i} = \frac{3}{2} \sum_{i = 0}^{\infty} \frac{i}{3^{i}} + \frac{1}{4} \sum_{i = 0}^{\infty} \frac{1}{3^{i}}

= \frac{3}{2} \frac{1}{4} + \frac{1}{4} \frac{3}{2} = \frac{3}{8} + \frac{3}{8} = \frac{6}{8} = \frac{3}{4} \Rightarrow S = \frac{3}{4} .