What does this double sided arrow $\longleftrightarrow$ mean? What is $\longleftrightarrow$ used for in mathematics? I know about $\iff$ being used for "If and only if". Are they the same thing? I was watching a YouTube video that said:
$$\sum^{\infty}_{n=1} {1\over n^x} \longleftrightarrow \int^{\infty}_{1} {1\over t^x} dt$$
The teacher mentions convergence/divergence, but I was confused when the notation came up.
 A: On the one hand, $\longleftrightarrow $ is used for connecting propositional formulas (e.g. $p\to q \lor (p\longleftrightarrow q) \land \lnot w$). You can understand it as a binary operator like AND or OR, which are represented by $\land $ and $\lor $ symbols, as you would know.
Here you can see its truth table.
$$\begin{array}{|c|c|c|}
\hline
p&q&p\longleftrightarrow q\\
\hline
T&T&T\\
\hline
T&F&F\\
\hline
F&T&F\\
\hline
F&F&T\\
\hline
\end{array}$$
On the other hand, $\iff $ is used as a connective of propositional formulas. You can see both uses here:
$$p\longleftrightarrow q \iff (p\to q) \land (q \to p)$$
And what does $a \iff b $ means? If you write $a\iff b $, then you could actually say the same by writing down that the bicondition $a \text { is true} \longleftrightarrow b \text{ is true} $ is always true. Note that this works whatever the truth values of $a \text { is true} $ or $b \text { is true}$ are.
Edit: in another fields a part of logic, (at least in basic degrees), choosing one or the other does not matter too much ($\longleftrightarrow $ or $\iff $ are just "lazy" math translations of simple English connector "if and only if").
A: As has been mentioned in the comments, this is almost certainly an idiosyncratic use, and the author (is that the right word for somebody who makes a YouTube video? Probably not) ought to have explained what he or she intended the symbol to mean.  Without any additional context, it's hard to know for sure, but I'm going to hazard a guess that the symbol is intended to denote "are equivalent" in some (perhaps ill-defined) sense.  In what sense?  Probably in the sense of "equiconvergence" -- i.e., their convergence behavior is equivalent (one of them converges if and only if the other one does).
A: I understand it, in this context, as a sign corresponding to a "loose equivalence" between the convergence of both terms, under specific conditions that are not  fully mentioned. The presenter writes that a sum diverges/converges if the corresponding integral  diverges/converges. I would translate it as: "LHS property is (somehow) strongly related to the RHS one".
It is not an "if and only if", indeed this is not true in general in that case.
The $\leftrightarrow$ symbol appears after the Maclaurin–Cauchy integral test for  convergence (the so-called Cauchy integral theorem is quite different). The standard test works under the following conditions:

*

*$f$ is continuous, defined on $[n_0, +\infty [$ for some integer $n_0$,

*$f$ is monotone and decreasing.

Then the infinite series $\sum_{n=n_0}^\infty f(n)$
converges to a finite limit if and only if ($\Leftrightarrow$) the improper integral
$\int_N^\infty f(x)\,dx$ is finite. And if the integral diverges, then the series diverges as well. Here, the test works for the $p$-series, as $t \to \frac{1}{t^x}$ is continuous decreasing for $x >0$, and the convergence of the series depends on $x> 1$ or not.
As mentioned in comments, many mathematical symbols have several interpretations (e.g. bijection or logical biconditional).
A: In the area of logic, $\longleftrightarrow$ is usually used for "if and only if" instead of $\iff$ (because who wants to bother drawing that second line all the time).
Otherwise when dealing with functions, $\longleftrightarrow$ might also be used to denote a bijective function. So $f \colon A \leftrightarrow B$ is a bijection between $A$ and $B$. Or you could similarly write
$$
A \overset{f}{\longleftrightarrow} B
$$
In regards to what was likely meant in the video that you saw, the following is true:
For a given value of $x$, one has $\sum\limits_{n=1}^\infty \frac{1}{n^x}$ converges if and only if $\int\limits_{1}^\infty \frac{1}{t^x}dt$ converges.
