What value does 1/sqrt(n) approach?

Question

Intuitively I would say 0 but I want to get a second opinion. We know that $\frac{1}{n}$ approaches 0, as n goes to infinity.

We also know that $\sum^\infty_{n=0}\frac{1}{n}$ diverges. We also know that $\sum^\infty_{n=0}\frac{1}{\sqrt{n}}$ diverges by the integral test and "direct comparison" test.

How can we conclude anything about the limit from this knowledge about the limit of $\frac{1}{\sqrt{n}}$ ?

When I plot it I can see it approaching zero but how can I proove it ?

Answer 1

Just use that $\sqrt{\cdot}$ is continuous at $0$ and $\lim_{n \to \infty} \frac{1}{n} = 0$. Then we get $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} = \sqrt{\lim_{n \to \infty} \frac{1}{n}} = \sqrt{0} = 0. $$

Answer 2

From the divergence of $\sum^\infty_{n=0}\frac{1}{\sqrt{n}}$ we can deduce nothing, but:

$\frac{1}{\sqrt{n}}=|\frac{1}{\sqrt{n}}-0|< \epsilon$ $\iff n > \frac{1}{\epsilon^2}$.

Can you now see that $\frac{1}{\sqrt{n}} \to 0$ for $n \to \infty$ ?