# Using the integral test, how do you show whether #(1/sqrt (n+1))# diverges or converges?

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To determine whether the series ( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}} ) converges or diverges using the integral test, we compare it to the integral of the corresponding function.

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

In this case, let ( f(x) = \frac{1}{\sqrt{x+1}} ).

Now, we check the convergence of the integral ( \int_{1}^{\infty} \frac{1}{\sqrt{x+1}} , dx ).

[ \int_{1}^{\infty} \frac{1}{\sqrt{x+1}} , dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{\sqrt{x+1}} , dx ]

[ = \lim_{b \to \infty} \left[2\sqrt{x+1}\right]_{1}^{b} ]

[ = \lim_{b \to \infty} \left[2\sqrt{b+1} - 2\sqrt{1+1}\right] ]

[ = \lim_{b \to \infty} \left[2\sqrt{b+1} - 2\right] ]

[ = \infty ]

Since the integral diverges, by the integral test, the series ( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}} ) also diverges.

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