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

Question

Christopher Allers · Accepted Answer

The sum is unbounded

We know

#sum_{i=1}^{n} 1 / [sqrt(i) * (sqrt(i) + 1)]geint_{1}^n(dx)/(sqrt(x)(sqrt(x)+1)} = 2log_e(1+sqrt(n))#

because # 1 / [sqrt(i) * (sqrt(i) + 1)]# is monotonic decreasing.

but #lim_{n->oo}2log_e(1+sqrt(n)) = oo#

so the sum is unbounded

Attached a figure with the comparisson between series and integral.

Charles Arocha · Answer

undefined 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.