# Series estimation/error

1. Nov 27, 2012

### APolaris

The question is: estimate the sum, from 2 to infinity, of 1 / (n^2 + 4) to within 0.1 of exact value.

I have the following: the integral from n to inf. of 1 / (x^2 + 4) is (pi/4 - 1/2 arctan(n/2)).

Next, in order to find the number of partial sums to use, set Sn + the integral from n+1 to infinity < S < Sn + the integral from n to infinity.

This gives (pi/4 - 1/2 arctan[(n+1)/2]) < (remainder) < (pi/4 - 1/2 arctan(n/2)).

I can only assume from here that you are meant to subtract the lesser from the greater and set (1/2 arctan[(n+1)/2]) - 1/2 arctan(n/2) < 0.2. So arctan((n+1)/2) - arctan(n/2) < 0.4. What?

Last edited: Nov 27, 2012
2. Nov 27, 2012

### slider142

The error of the integral approximation will be between the two integrals you have, so you want
$$\int_{n+1}^\infty \frac{dx}{x^2 + 4} \leq 0.1 \leq \int_n^\infty \frac{dx}{x^2 + 4}$$
These two separate inequalities will tell you what range of values of n you should choose to approximate the series within an error bound of 0.1. For this problem, it turns out there is only one integer between these two integrals.
$$\frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{n + 1}{2}\right) \leq 0.1$$
for example gives us the lower bound when solved of
$$n \geq 2\tan\left(\frac{\pi}{2}-0.2\right)-1 \approx 8.86631$$
Once you have the value of n, you can use it in the inequality
$$s_n + \int_{n+1}^\infty \frac{dx}{x^2 + 4} \leq s \leq s_n + \int_n^\infty \frac{dx}{x^2 + 4}$$
Adding the two values and dividing by 2 does not change the error of the value, so you can approximate the value s by taking the average of the two values outside.

Last edited: Nov 27, 2012