For example, does the following summation "converge" to $\infty$?
$$\sum_{n=1}^{\infty}\frac 1 n$$
If so, give me some other examples and explain what it means to converge to infinity.
Also, is there anything special about a series converging to infinity?
One says that $\displaystyle\sum_{n=0}^\infty \frac 1 {2^n}$ "converges to $2$", NOT that it "converges on $2$".
One says that $\displaystyle\sum_{n=0}^\infty (-1)^n$ "diverges", but that does not mean it "diverges to" anything.
But with $\displaystyle\sum_{n=1}^\infty \frac 1 n$, one could say that it "converges to $\infty$" for the same reason one says the first sum above "converges to $2$"; however, it is conventional to say that it "diverges to $\infty$".
I suspect that ultimately the justification of saying "diverges to $\infty$" rather than "converges to $\infty$" is that $\infty$ is an absorbent element for addition and multiplication: If you add any finite number to $\infty$ or multiply any finite number by $\infty$, you get $\infty$, and this matters if you consider things like $\displaystyle\sum_n \left( \frac 1 n + \frac 1 {2^n} \right)$, etc. That would explain why infinite products $\displaystyle\prod_{n=1}^\infty a_n$ of positive numbers $a_n$ are in some cases said to "diverge to $0$" rather than to "converge to $0$". They many "converge to" any positive real number, but if the limit is $0$ or $\infty$ they are said to "diverge", just as if you had something oscillating like $\displaystyle N\mapsto\sum_{n=1}^N (-1)^n$.
This is one case where the more general definition of a limit is handy:
A sequence $s_n$ converges to $L$ if, for any neighborhood of $L$, there is some $N$ such that all the $s_m$ with $m>N$ are within this neighborhood.
Where the word "neighborhood" refers to some open set containing $L$. In particular, if you take $(L-\varepsilon,L+\varepsilon)$ to be a neighborhood you recover the $\varepsilon-\delta$ definition of a limit.
Following this, one can see that the series $\sum_{n=1}^{\infty}\frac{1}n$ diverges in $\mathbb R$ because there is no such $L$. The easiest way to prove that is to see that it exhibits unbounded growth. However, one can also work in the extended real line where we add $\infty$ and $-\infty$ to our set and say that a neighborhood of $\infty$ looks like a set $(c,\infty]$ - meaning to converge to $\infty$ the sequence must grow beyond any finite bound. In this sense, you can find that $\sum_{n=1}^{\infty}\frac{1}n$ does indeed converge to $\infty$.
So, whether a series converges or diverges is somewhat dependent on the space in which it exists - "converges to $\infty$" basically means "grows without bound" and is a natural notion when we include $\infty$ in our set.
Divergence is based on the end of the series
do the terms settle on a number (Converges)
do they go to ±∞ (Diverges)
do they bounce between different numbers without settling on one specifically (Diverges)
not on how you get there
The reason we don't say that a series converges to infinity but rather *diverges" to infinity is so as to respect the equivalence between "convergent" and "Cauchy". It is true that in some ways for a series diverging to infinity, intuitively looks like the series of partial sums is converging, but at any rate it is not a Cauchy sequence.
In more detail, a sequence $(u_n)$ is Cauchy if, as Cauchy defined it, the difference $u_n-u_m$ is infinitesimal for all infinite indices $n$ and $n$. Meanwhile, a sequence $(u_n)$ that tends to infinity will not have this property. For example for the sequence with $n$-th term $u_n=\sqrt{n}$, the consecutive terms $u_n$ and $u_{n+1}$ are getting closer and closer, but for example $u_n$ and $u_{2n}$ are not getting closer when $n$ tends to infinity.
To be a Cauchy sequence, $|x_m-x_n|$ must be infinitesimal when m and n are infinite. This happens to be the case for $\sum_{n=1}^{\infty}\frac 1 n$. So does it converge?
– Simply Beautiful Art Nov 30 '15 at 23:56