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How calculate `(1+1/2+ ... +1/2^n)(1+1/3+ ... +1/3^n)` if n tends to infinity?

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markll | Student, College Freshman | eNoter

Posted September 8, 2012 at 4:42 PM via web

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How calculate `(1+1/2+ ... +1/2^n)(1+1/3+ ... +1/3^n)` if n tends to infinity?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 8, 2012 at 5:01 PM (Answer #1)

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The value of `(1+1/2+ ... +1/2^n)(1+1/3+ .... +1/3^n)` has to be determined for n tending to infinity.

`1+1/2+ ... +1/2^n` is the sum of a geometric series with common denominator (1/2). The sum of infinite terms of this series is `1/(1 - 1/2) = 1/(1/2) = 2`

Similarly `1+1/3+ ... +1/3^n = 1/(1 - 1/3) = 1/(2/3) = 3/2`

The product `(1+1/2+...+1/2^n)(1+1/3+ ... +1/3^n) = 2*(3/2) = 3`

The required value of `(1+1/2+ ... +1/2^n)(1+1/3+ ... +1/3^n)` if n tends to infinity is 3

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