Using the integral test, how do you show whether #sum 1/n^2# diverges or converges from n=1 to infinity?

Question

Paisley Johnson · Accepted Answer

Since #sum_1^(+oo)a_n<=int_1^(+oo)a_xdx#, if the integral is convergent, so it will be the series. Since #int_1^(+oo)1/x^2dx# is convergent because it's like: #int_1^(+oo)1/x^alphadx# with #alpha>1#, it is convergent also the series. If you want, you can do the integral, but it is useless! #int_1^(+oo)1/x^2dx=[x^(-2+1)/(-2+1)]_1^(+oo)=[-1/x]_1^(+oo)=0+1=1#.

Christopher Allers · Answer

undefined To determine whether the series $ \sum_{n=1}^{\infty} \frac{1}{n^2} $ converges or diverges, we can use the integral test.

The integral test states that if $ f(x) $ is a continuous, positive, and decreasing function for $ x \geq 1 $ and $ f(n) = a_n $ for all positive integers $ n $, then the series $ \sum_{n=1}^{\infty} a_n $ converges if and only if the improper integral $ \int_{1}^{\infty} f(x) \, dx $ converges.

For $ f(x) = \frac{1}{x^2} $, let's evaluate the improper integral:

$$ \int_{1}^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^2} \, dx $$

$$ = \lim_{t \to \infty} \left[ -\frac{1}{x} \right]_{1}^{t} $$

$$ = \lim_{t \to \infty} \left( -\frac{1}{t} + \frac{1}{1} \right) $$

$$ = \lim_{t \to \infty} \left( 1 - \frac{1}{t} \right) $$

$$ = 1 $$

Since the integral $ \int_{1}^{\infty} \frac{1}{x^2} \, dx $ converges, by the integral test, the series $ \sum_{n=1}^{\infty} \frac{1}{n^2} $ also converges.