Question

`lim_(x rightarrow oo) (n^2/((n^2 + 1)(n + 1)) + n^2/((n^2 + 4)(n + 2)) + n^2/((n^2 + 9)(n + 3)) + ... + n^2/((n^2 + n^2)(n + n)))` is equal to ______.

पर्याय

• `π/8 + 1/4 log_e 2`

• `π/4 + 1/8 log_e 2`

• `π/4 - 1/8 log_e 2`

• `π/8 + log_e sqrt(2)`

Answer 1

`lim_(x rightarrow ∞) (n^2/((n^2 + 1)(n + 1)) + n^2/((n^2 + 4)(n + 2)) + n^2/((n^2 + 9)(n + 3)) + ... + n^2/((n^2 + n^2)(n + n)))` is equal to `underlinebb(π/8 + 1/4 log_e 2)`.

Explanation:

Given expression is `lim_(n rightarrow ∞) (sum_(r = 1)^n n^2/((n^2 + r^2)(n + r)))`

= `lim_(n rightarrow ∞) (sum_(r = 1)^n 1/(n(1 + (r/n)^2)(1 + (r/n))))`

Here, `(r/n)` = x, `(1/n)` = dx

= `int_0^1 (dx)/((1 + x^2)(1 + x))`

= `1/2 int_0^1 (1 - x)/(1 + x^2) dx + 1/2 int_0^1 1/(1 + x)dx`

= `1/2 int_0^1 (1/((1 + x^2)) - x/((1 + x^2)))dx + 1/2 (ln(1 + x))_0^1`

 = `1/2[tan^-1x - 1/2 ln (1 + x^2)]_0^1 + 1/2 ln 2`

Apply the limit,

= `1/2[π/4 - 1/2 ln 2] + 1/2 ln 2`

= `π/8 + 1/4 ln 2`