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

Since the sum $\sum_{k=1}^\infty \frac{1}{k^2}$ is convergent, you can make the "tail" of the sum ($\sum_{k=n+1}^\infty \frac{1}{k^2}$) as small as you want.
Explicitly, since $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$, for any $\epsilon>0$, you can find an $n$ such that $\sum_{k=n+1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}-\sum_{k=1}^n \frac{1}{k^2}<\epsilon$.
In that case, the tail sums form a strictly decreasing sequence of positive numbers, which can be made smaller than any $\epsilon>0$.
That is convergence to 0! :)
Naturally if the expression converges to zero, so does the square root of the expression. 
