# Given function fn(x)=1/(n^2+x^2), show that limit n(fn(1)+fn(2)+---fn(n)) = `pi` /4 if n natural number?

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### 1 Answer

You need to check if evaluating the limit `lim_(n->oo) n(fn(1) + fn(2) + ... + fn(n)) = pi/4` , hence, you need first evaluate the summation `fn(1) + fn(2) + ... + fn(n)` , such that:

`fn(1) + fn(2) + ... + fn(n) = 1/(1^2+n^2) + 1/(2^2+n^2) + ... + 1/(n^2+n^2)`

Factoring out `1/n^2` yields:

`fn(1) + fn(2) + ... + fn(n) = (1/n^2)(1/(1 + (1/n)^2) + 1/(1 + (2/n)^2) + ... + 1/(1 + (n/n)^2))`

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

`n(fn(1) + fn(2) + ... + fn(n)) = (1/n)(1/(1 + (1/n)^2) + 1/(1 + (2/n)^2) + ... + 1/(1 + (n/n)^2))`

`n(fn(1) + fn(2) + ... + fn(n)) = (1/n) sum_(k=1)^n 1/(1 + (k/n)^2)`

You need to notice that the summation `n(fn(1) + fn(2) + ... + fn(n))` represents the Riemann summation associated to the function `g(x) = 1/(1 + x^2),` hence, you may evaluate the limit, such that:

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

`lim_(n->oo) n(fn(1) + fn(2) + ... + fn(n)) = tan^(-1) x|_0^1`

By fundamental theorem of calculus, yields:

`lim_(n->oo) n(fn(1) + fn(2) + ... + fn(n)) = tan^(-1) 1 - tan^(-1) 0`

`lim_(n->oo) n(fn(1) + fn(2) + ... + fn(n)) = pi/4 - 0`

`lim_(n->oo) n(fn(1) + fn(2) + ... + fn(n)) = pi/4`

**Hence, checking if evaluating the limit `lim_(n->oo) n(fn(1) + fn(2) + ... + fn(n))` yields `pi/4` , the statement is valid.**

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