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: 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 $N \rightarrow \infty$.
A: 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.
A: In general we have if $\sum_{k=1}^{\infty}a_k$ converges then the tail of the sereis $\sum_{k=N}^{\infty}a_k$must go to zero to see this consider $$s_n=\displaystyle\sum_{k=1}^{n}a_k$$ then we have$s_n\to\sum_{k=1}^{\infty}a_k$ that means for any $\epsilon$ we there is $N$ such that$$|\displaystyle\sum_{k=1}^{\infty}a_k-s_N|<\epsilon$$ but $$|\displaystyle\sum_{k=1}^{\infty}a_k-s_N|=|\displaystyle\sum_{k=N}^{\infty}a_k|$$ thus, $$|\displaystyle\sum_{k=N}^{\infty}a_k|<\epsilon$$ since epsilon was arbitrary we conclude that$\sum_{k=N}^{\infty}a_k\to 0$. as for your example we know by the integral test that $\sum_{k=1}^{\infty}\frac{1}{k^2}$ is convergent Thus, by what I wrote we must have $\sum_{k=N}^{\infty}\frac{1}{k^2}\to 0.$ Remark: of course if $\sum_{k=N}^{\infty}\frac{1}{k^2}\to 0.$ then $(\sum_{k=N}^{\infty}\frac{1}{k^2})^{1/2}\to 0.$
A: 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 $\sum_1^\infty \frac{1}{n^2}$.
A: 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. 
A: Notice that $\sum \frac{1}{k^2}$ is convergent.  If I invoke Cauchy's Convergence Criterion, the proof of this sum is trivial.
A: 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 $\lim f(a_n) = f( \lim a_n )$.   
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{%
\pars{\sum_{k = n + 1}^{\infty}\ {1 \over k^{2}}}^{1/2}} =
\pars{\sum_{k = 1}^{\infty}\ {1 \over k^{2}} -
\sum_{k = 1}^{n}\ {1 \over k^{2}}}^{1/2}
\\[5mm] = &\
\braces{\zeta\pars{2} -
\bracks{-{1 \over n} + \zeta\pars{2} + 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\dd x}}^{1/2}
\end{align}
where I used a zeta function identity. Moreover,
$$
0 < 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\dd x <
2\int_{n}^{\infty}{\dd x \over x^{3}} = {1 \over n^{2}}
$$
Then,
\begin{align}
&\bbox[5px,#ffd]{%
\pars{\sum_{k = n + 1}^{\infty}\ {1 \over k^{2}}}^{1/2}} =
\bracks{{1 \over n} - 2\int_{n}^{\infty}{\braces{x} \over x^{3}}\dd x}^{1/2}
\\[5mm] &\
\bbx{\to \color{red}{\large 0}\ \mbox{as}\ n \to \infty}
\\ &
\end{align}
