# How do I calculate : `lim [1*3*5*7....*(2n-1)]/ [2*4*6*8...*(2n)]` `n-gtoo`

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### 1 Answer

You may evaluate the limit using the principle of mathematical induction to prove the following inequality, such that:

`(1*3*5*...*(2n-1))/(2*4*...*2n) < 1/(sqrt(2n+1))`

The process of mathematical induction consists of two parts.

The first part, called the base case, proves that the inequality holds for n = 1, such that:

`1/2 < 1/(sqrt(2*1 + 2)) => 1/2 < 1/sqrt 3`

Since `2 > sqrt 3 => 1/2 < 1/sqrt 3` holds

The second part of the process, called the inductive step, proves that if the inequality holds for some number n, then it holds for n+1, such that:

`(1*3*5*...*(2n-1))/(2*4*...*2n) < 1/(sqrt(2n+1))` holds

`(1*3*5*...*(2(n+1)-1))/(2*4*...*2(n+1)) < 1/(sqrt(2(n+1)+1))`

`(1*3*5*...*(2n+1))/(2*4*...*(2n+2)) < 1/(sqrt(2n+3))`

ButÂ `(1*3*5*...*(2n+1))/(2*4*...*(2n+2)) = (1*3*5*...*(2n-1))/(2*4*...*2n)*(2n(2n+1))/((2n+1)(2n+2))`

Reducing duplicate terms yields:

`(1*3*5*...*(2n-1))/(2*4*...*2n)*(2n/(2n+2)) < 1/(sqrt(2n+1))*(n/(n+1)) < 1/(sqrt(2n+3)) => n/(sqrt(2n+1)) < (n+1)/(sqrt(2n+3))`

Since `n/(sqrt(2n+1)) < n/(sqrt(2n+3)) => n/(sqrt(2n+1)) < n/(sqrt(2n+3)) + 1/(sqrt(2n+3))` , hence, the inequality `(1*3*5*...*(2n+1))/(2*4*...*(2n+2)) < 1/(sqrt(2n+3))` holds.

Since both steps of mathematical induction have been proved, then, the inequality `(1*3*5*...*(2n-1))/(2*4*...*2n) < 1/(sqrt(2n+1))` holds.

You need to remember the property of limits of sequences and subsequences, such that:

`(x_n)<(y_n) => lim_(n->oo) x_n < lim_(n->oo) (y_n) => if lim_(n->oo) (y_n) = l => lim_(n->oo) x_n = l`

Considering the subsequence `(x_n) = (1*3*5*...*(2n-1))/(2*4*...*2n)` and the sequence `(y_n) = 1/(sqrt(2n+1)),` then, reasoning by analogy, yields:

`lim_(n->oo) (1*3*5*...*(2n-1))/(2*4*...*2n) = lim_(n->oo) 1/(sqrt(2n+1))`

Evaluating the limit `lim_(n->oo) 1/(sqrt(2n+1))` yields:

`lim_(n->oo) 1/(sqrt(2n+1)) = lim_(n->oo) 1/(sqrt lim_(n->oo) (2n+1)) = 1/(sqrtoo) = 0`

**Hence, evaluating the requested limit, using the inequalities and the principle of mathematical induction, yields **`lim_(n->oo) (1*3*5*...*(2n-1))/(2*4*...*2n) = 0.`