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

If you want, you can do the integral, but it is useless!

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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.

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