# Using the integral test, how do you show whether #sum 1 / [sqrt(n) * (sqrt(n) + 1)]# diverges or converges from n=1 to infinity?

The sum is unbounded

We know

because

but

so the sum is unbounded

Attached a figure with the comparisson between series and integral.

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We can use the integral test to determine whether the series ( \sum \frac{1}{\sqrt{n}(\sqrt{n}+1)} ) converges or diverges. The integral test states that if ( f(x) ) is continuous, positive, and decreasing for all ( x \geq 1 ), and if ( \int_{1}^{\infty} f(x) , dx ) converges, then the series ( \sum_{n=1}^{\infty} f(n) ) also converges; conversely, if the integral diverges, then the series diverges as well.

Let's evaluate the integral ( \int_{1}^{\infty} \frac{1}{\sqrt{x}(\sqrt{x}+1)} , dx ) to determine the convergence or divergence of the series.

We can start by making a substitution: let ( u = \sqrt{x} ), then ( du = \frac{1}{2\sqrt{x}} , dx ). Substitute these into the integral:

[ \int_{1}^{\infty} \frac{1}{\sqrt{x}(\sqrt{x}+1)} , dx = 2\int_{1}^{\infty} \frac{1}{u(u+1)} , du ]

Now perform partial fraction decomposition on the integrand:

[ \frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} ] [ A(u+1) + Bu = 1 ] [ (A+B)u + A = 1 ]

Comparing coefficients, we get ( A+B = 0 ) and ( A = 1 ), so ( B = -1 ).

Now integrate the decomposed fractions:

[ 2\int_{1}^{\infty} \left( \frac{1}{u} - \frac{1}{u+1} \right) , du = 2\left[ \ln(u) - \ln(u+1) \right]_{1}^{\infty} ]

As ( u ) approaches infinity, both ( \ln(u) ) and ( \ln(u+1) ) approach infinity, but their difference remains bounded. Therefore, the integral converges.

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

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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