How to evaluate 1(3) 3(5) 5(7) … (2n-1)(2n 1)

Sequence sums by Archimedean property: lim(n→∞)⁡(3n+1)/(2n+5)=3/2 and lim(n→∞)⁡(n2-1)/(n3+3)=1/2

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The value of `i^(1+3+5+...+(2n+1))` is (1) i if n is even, -i if n is odd (2) 1 if n is even, ...

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Find the Sum of the Series SUM((2^n + 1)/3^n)

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For each `n in N, 3.(5^(2n+1))+2^(3n+1)` is divisible by

`1^2/(1.3)+2^2/(3.5)+3^2/(5.7)+.....+n^2/((2n-1)(2n+1))=((n)(n+1))/((2(2n+1))`

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If u(n)=(2n+1)/(3n+5) show that lim un = 2/3 by definition of limit.

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