If-a-lt-1-and-b-lt-1-then-the-sum-of-the-series-a-a-b-a-2-a-2-b-2-a-3-a-3-b-3-upto-is-

Question Number 43766 by peter frank last updated on 15/Sep/18

If ∣ a ∣< 1 and ∣ b ∣< 1, then the sum of the

series a (a + b) + a^{2} (a^{2} + b^{2}) + a^{3} (a^{3} + b^{3}) + \dots

upto \infty is

Answered by MrW3 last updated on 15/Sep/18

a (a + b) + a^{2} (a^{2} + b^{2}) + a^{3} (a^{3} + b^{3}) + \dots

= (a^{2} + a^{4} + a^{6} + \dots) + [{(a b)}^{2} + {(a b)}^{4} + {(a b)}^{6} + \dots]

\to \frac{a^{2}}{1 - a^{2}} + \frac{{(a b)}^{2}}{1 - {(a b)}^{2}}

= \frac{a^{2} (1 - a^{4} b^{2})}{(1 - a^{2}) (1 - a^{2} b^{2})}

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