Why does $\sum \frac{1}{n^{1 + \epsilon}}$ converge? The proof that the infinite sum of $\frac{1}{n}$ diverges seems to have a fair amount of breathing room.  We group successive terms in quantities of increasing powers, starting with $\frac{1}{2}$, then $\frac{1}{3} + \frac{1}{4}$, then the next four terms, then the next eight terms, etc., and note that each of the groups is greater than or equal to $\frac{1}{2}$, and adding $\frac{1}{2}$ forever approaches $\infty$.
As extra credit for this proof, each group after the first is strictly greater than $\frac{1}{2}$, so the divergence actually occurs faster.  Furthermore, we didn't even need the terms to be that large; adding $\frac{1}{1,000,000}$ forever would also approach $\infty$.  Why then, given this generous cushion in the proof, is it the case that $\frac{1}{n^{1 + ε}}$ for some tiny ε converges?  Why is the power of $n$ so fragile to nudges in the positive direction given how firmly $\frac{1}{n}$ seemed to diverge?
 A: I will address the last paragraph of your post. As long as epsilon is a positive fixed quantity, the series will converge. This can be seen with the Integral test. $1/x$ integrates to $lnx$ and with $x$ going to infinity, the integral and thus the series diverges. But if the exponent is more than $1$, the polynomial term integrates to another polynomial term. I leave it up to you the figure out why that implies convergence because this essentially will answer your last part of the question. Lastly (and not unimportant!) a note on the word "fixed". If the epsilon is not a fixed positive quantity, but variable, then the series may be divergent. For example the series $\frac{1}{n^{(1+1/n)}}$ has a "variable" exponent, but the exponent is more than $1$. Yet this series turns out to be divergent. 
A: Also, the integral test
can be used to show that
$\sum \dfrac1{n\ln \ln ... \ln(n)}
$
diverges for
any fixed number of
nexted $\ln$.
This is because,
if we define
$\ln_0(n) = 1
$
and
$\ln_{k+1}(n)
=\ln(\ln_k(n))
$,
then
$(\ln_k(x))'
=\dfrac1{x\prod_{j=1}^{k-1}\ln_{j}(x)}
$.
Proof.
$(\ln_1(x))'
=(\ln(x))'
=\dfrac1{x}
$
and
$(\ln_2(x))'
=(\ln(\ln(x)))'
=(\ln(x))'\dfrac1{\ln(x)}
=\dfrac1{x\ln(x)}
$
If
$(\ln_k(x))'
=\dfrac1{x\prod_{j=1}^{k-1}\ln_{j}(x)}
$,
then
$\begin{array}\\
(\ln_{k+1}(x))'
&=(\ln(\ln_k(x)))'\\
&=(\ln_k(x))'\dfrac1{\ln_k(x)}\\
&=\dfrac1{x\prod_{j=1}^{k-1}\ln_{j}(x)\ln_k(x)}\\
&=\dfrac1{x\prod_{j=1}^{k}\ln_{j}(x)}\\
\end{array}
$
Since
$=\ln_k(x)
\to \infty$
as $x \to \infty$
for any fixed $k$,
$\int \dfrac{dx}{x\prod_{j=1}^{k-1}\ln_{j}(x)}
=\ln_k(x)
\to \infty$
as $x \to \infty$
so
$\sum \dfrac1{n\prod_{j=1}^{k-1}\ln_{j}(n)}
$
diverges by the integral test.
You can similarly show that
$\sum \dfrac1{n\prod_{j=1}^{k-1}\ln_{j}(n)\ln_{k}^{1+\epsilon}(n)}
$
converges for any fixed $k$
and $\epsilon > 0$.
A: To address the question why a small $\epsilon$ is enough, note that analogously to the proof of divergence for the harmonic series, we can say that
$$\frac1{2^{1+\epsilon}} \ge \frac1{2^{1+\epsilon}}, $$
$$\frac1{3^{1+\epsilon}} + \frac1{4^{1+\epsilon}} \ge 2\frac1{4^{1+\epsilon}} = \frac1{2^{1+2\epsilon}}, $$
$$\frac1{5^{1+\epsilon}} + \frac1{6^{1+\epsilon}} + \frac1{7^{1+\epsilon}} + \frac1{8^{1+\epsilon}}\ge 4\frac1{8^{1+\epsilon}} = \frac1{2^{1+3\epsilon}}, $$
and so on. Note that the terms on the right hand side are no longer are constant as they used to be for the harmonic series, instead they form a geometric progression with the factor $\frac1{2^\epsilon}$. When $\epsilon$ is small, that value is just a tiny bit smaller than $1$. 
Still any geometric sequence with factor $< 1$ will converge to $0$, even if slowly. That means if we take the sum of the right hand sides, it is no longer an infinite sum of $\frac12$ that diverges, but a geometric series that does converge!
So the main "problem" in translating the proof for $\epsilon>0$ is that our divering minorante for the harmonic series no longer divererges!
In essence it is just the difference that $\sum_{i=0}^{\infty}\frac12$ diverges, while $\sum_{i=0}^{\infty}\frac1{2^{1+i\epsilon}}$ converges. 
This insight allows you to actually prove that $\sum_{n=0}^{\infty}\frac1{n^{1+\epsilon}}$ converges without the integral methods mentioned in other answers.
That's because 
$$\frac1{3^{1+\epsilon}} + \frac1{4^{1+\epsilon}} \le 2\frac1{2^{1+\epsilon}} = \frac1{2^{\epsilon}}, $$
$$\frac1{5^{1+\epsilon}} + \frac1{6^{1+\epsilon}} + \frac1{7^{1+\epsilon}} + \frac1{8^{1+\epsilon}}\le 4\frac1{4^{1+\epsilon}} = \frac1{2^{2\epsilon}}, $$
$$\frac1{9^{1+\epsilon}} + \frac1{10^{1+\epsilon}} + \frac1{11^{1+\epsilon}} + \frac1{12^{1+\epsilon}} +\frac1{13^{1+\epsilon}} + \frac1{14^{1+\epsilon}} + \frac1{15^{1+\epsilon}} + \frac1{16^{1+\epsilon}} \le 8\frac1{8^{1+\epsilon}} = \frac1{2^{3\epsilon}}, $$
a.s.o. 
Now our series has a converging majorante $\sum_{i=0}^{\infty}\frac1{2^{i\epsilon}}$, so converges itself.
