Logical Biconditional Doubt (Applied to a Case)

Question

The biconditional logical connective (⇔) produces a value of true if and only if both operands are false or both operands are true. Now, I am studying Calculus II improper integrals. Specifically, the limit comparison test: Be, $f(x)≥0$ and $g(x)>0$ within $[a;+∞)$. And be $f(x), g(x)$ Riemann integrable functions within $[a;b]$, $∀b≥a$.

And be it, $$\lim _{x\to +\infty }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=L$$

Then, if $0<L<∞$

$$\int _a^{+∞}g\left(x\right)dx\:converges\:⇔\int _a^{+∞}f\left(x\right)dx\:converges\:$$

Does that mean that if one of them diverges, then the other is going to diverge as well? It seems intuitively logical to me but I urge to be certain of this.

Yes, that is correct. You do understand the biconditional. (But my spell checker does not.) — GEdgar, Mar 19 '20 at 16:44
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score 1 · Accepted Answer · answered Jun 09 '20 at 03:14

The biconditional ($\iff$) is a logical connective that is true when both operands are true, or both are false $(0)$, it is false otherwise.

$$(\color{red}{\lnot A}\land\color{red}{\lnot B})\lor(\color{blue}{A}\land \color{blue}{B})\tag{0}$$

The truth table of the biconditional is the following:

\begin{array}{c|cc} &\lnot A&A\\\hline \lnot B&\color{red}{1}&0\\ B&0&\color{blue}{1} \end{array}

The bicontional of two statements is equivalent to the conjunction of two if ($\impliedby$, $\implies$) statements $(1)$, meaning that the truth of the LHS of a biconditional, implies the truth of the RHS.

$$A\iff B \equiv (A\impliedby B)\land(A\implies B)\tag{1}$$

The biconditional can be referred to as if and only if, which is often abbreviated to iff. In $(1)$, it can be said that the truth of $A$, is a necessary and sufficient condition for the truth of $B$, which is another way to refer to the biconditional.

In your example, the assertion $(2)$ states that $\int _a^{+∞}g\left(x\right)dx$ will converge if and only if $\int_a^{+\infty}f\left(x\right)dx$ does. Since divergence is the negation of convergence, $(2)$ also states that the $1^{st}$ integral will diverge iff the $2^{nd}$ does as well.

$$\int _a^{+\infty}g\left(x\right)dx\:\text{converges}\:\iff\int _a^{+ \infty}f\left(x\right)dx\:\text{converges}\:\tag{2}$$

Logical Biconditional Doubt (Applied to a Case)

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