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?
$$\lim_{n \to \infty}\left ( \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2} + \frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$=\left (\lim_{n \to \infty}\frac{1}{(n+1)^2}+ \lim_{n \to \infty}\frac{1}{(n+2)^2} + \lim_{n \to \infty}\frac{1}{(n+3)^2} + \dots \right)^{1/2} \to 0$$
$$(0 + 0 + 0 + \dots)^{1/2} = 0$$

A related problem. Note that,
$$ \sum_{k=n+1}^{\infty} \frac{1}{k^2}= \sum_{k=1}^{\infty} \frac{1}{(k+n)^2} \leq \sum_{k=1}^{\infty} \frac{1}{k^2} <\infty, $$
which implies that the series 
$$ \sum_{k=1}^{\infty} \frac{1}{(k+n)^2} $$
converges uniformly. So, we have
$$ \lim_{n\to \infty} \sqrt{  \sum_{k=n+1}^{\infty}\frac{1}{k^2}  } = \sqrt{\lim_{n\to \infty}  \sum_{k=n+1}^{\infty}\frac{1}{k^2}  } = \sqrt{\lim_{n\to \infty}  \sum_{k=1}^{\infty}\frac{1}{(k+n)^2}  } = \sqrt{  \sum_{k=1}^{\infty}\lim_{n\to \infty}\frac{1}{(k+n)^2} }=0. $$
Notice that, the function $\sqrt{x}$ is a continuous function for $x>0$, which justifies changing the operation $\lim f(a_n) = f( \lim a_n )$.   
