<s>[INST] How does this sum go to $0$? http://www.math.chalmers.se/Math/Grundutb/CTH/tma401/0304/handinsolutions.pdf
In problem (2), at the very end it says 
$$\left(\sum_{k = n+1}^{\infty} \frac{1}{k^2}\right)^{1/2} \to 0$$
I don't see how that is accomplished. I understand the sequence might, but how does the sum $$\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$?
Is one allowed to do the following?
$$\lim_{n \to \infty}\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$=\left (\lim_{n \to \infty}\frac{1}{(n+1)^2}+ \lim_{n \to \infty}\frac{1}{(n+2)^2} + \lim_{n \to \infty}\frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$(0 + 0 + 0 + \dots)^{1/2} = 0$$
 [/INST] 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 $\sum_1^\infty \frac{1}{n^2}$.
</s>