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)).`

**Further Reading**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now