How does this sum go to $0$?

One can forget about the square root part for a while. Note that $\dfrac{1}{k^2}\lt \dfrac{1}{(k-1)k}$.

But $\dfrac{1}{(k-1)k}=\dfrac{1}{k-1}-\dfrac{1}{k}$. Thus your sum is less than $$\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}+\cdots.$$ Note the wholesale cancellation: the above sum is $\dfrac{1}{n}$.

It follows that your original expression is less than $\dfrac{1}{\sqrt{n}}$.

Remark: Treating infinite "sums" as if they were long finite sums is a dangerous business that can all too easily give wrong answers. If one has experience with a particular series, such as the convergent series $\sum_1^\infty \frac{1}{n^2}$, then one can "see" that the tail must approach $0$. In fact, the issue is precisely the issue of the convergence of $\sum_1^\infty \frac{1}{n^2}$.

It turns out that

$$\sum_{k=N+1}^{\infty} \frac{1}{k^2} = \int_0^{\infty} dx \frac{x }{e^x-1} e^{-N x}$$

You can prove this by factoring out an $e^{x}$ from the denominator and Taylor expanding the resulting denominator. In any case, by integrating by parts, you can show that

$$\sum_{k=N+1}^{\infty} \frac{1}{k^2} = \frac{1}{N} + O\left( \frac{1}{N^2}\right)$$

In fact, you can get a full asymptotic expansion of the sum over $N$ using this integral. In any case, though, this shows how the sum vanishes as $N \rightarrow \infty$.

First note that $$\dfrac1{k^2} \leq \int_{k-1}^k\dfrac{dx}{x^2} \,\,\,\, (\text{Why?})$$ Hence, $$\sum_{k=n+1}^{\infty} \dfrac1{k^2} \leq \sum_{k=n+1}^{\infty} \int_{k-1}^k\dfrac{dx}{x^2} = \int_n ^{\infty} \dfrac{dx}{x^2} = \dfrac1n$$ Hence, we have that $$0 \leq \lim_{n \to \infty} \sum_{k=n+1}^{\infty} \dfrac1{k^2} \leq \lim_{n \to \infty} \dfrac1n = 0$$ Your argument to prove it is incorrect, since in general you cannot swap two limits to conclude the answer. For instance, by your same argument, you will also get that $$\lim_{n \to \infty} \left(\dfrac1{n} + \dfrac1{n} + \cdots + \dfrac1n + \cdots \right) = 0$$which is clearly false.