Calculate limit of series an = n(1/(4n^2+1) + 1/(4n^2+4)+---+1/(5n^2))?

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You need to factor out `1/n^2` , such that:

`a_n = n*(1/n^2)*(1/(4 + (1/n)^2) + 1/(4 + (2/n)^2) + ... + 1/(4 + (n/n)^2))`

Reducing duplicate factors yields:

`a_n = 1/n*(1/(4 + (1/n)^2) + 1/(4 + (2/n)^2) + ... + 1/(4 + (n/n)^2))`

`a_n = 1/n*sum_(k=1)^n 1/(4 + (k/n)^2)`

You should notice that a_n represents the Riemann summation associated to the function `f(x) = 1/(4 + x^2),` hence, you may evaluate the limit, such that:

`lim_(n->oo) a_n = int_0^1 1/(4 + x^2) dx`

`int_0^1 1/(4 + x^2) dx = (1/2) tan^(-1)(x/2)|_0^1`

`int_0^1 1/(4 + x^2) dx = (1/2) (tan^(-1)(1/2) - tan^(-1) 0)`

`int_0^1 1/(4 + x^2) dx = (1/2) (tan^(-1)(1/2))`

Hence, evaluating the limit of the given sequence a_n, using Riemann smmation, yields `lim_(n->oo) a_n = (1/2) (tan^(-1)(1/2)).`

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