Question

`lim_(n→∞){(1 + 1/n^2)^(2/n^2)(1 + 2^2/n^2)^(4/n^2)(1 + 3^2/n^2)^(6/n^2) ...(1 + n^2/n^2)^((2n)/n^2)}` is equal to ______.

Options

• `e/4`

• `4/e`

• 1

• `e/2`

Answer 1

`lim_(n→∞){(1 + 1/n^2)^(2/n^2)(1 + 2^2/n^2)^(4/n^2)(1 + 3^2/n^2)^(6/n^2) ...(1 + n^2/n^2)^((2n)/n^2)}` is equal to `underlinebb(4/e)`.

Explanation:

Taking logarithm both sides we get,

log_s = `2/n^2log(1 + 1^2/n^2) + 4/n^2log(1 + 2^2/n^2) + ..... + (2n)/n^2log(1 + n^2/n^2)`

= `sum_(r=1)^n(2r)/n^2(1 + r^2/n^2) = int_0^1 2xlog(1 + x^2)dx`

Put 1 + x² = t

⇒ 2xdx = dt

= `int_1^2logtdt = [tlogt - t]_1^2` = 2log2 – 2 + 1

logs = log4 – loge

⇒ s = `4/e`