Is every sufficient condition also a necessary condition?

Question

I've been studying calculus and I understand what sufficient and necessary conditions are, but is it true that every sufficient condition is also a necessary one?

For example:

Let $\sum{a_n}$ be a convergent series. If the series is convergent, then $a_n\to 0$. From this we can conclude that necessary condition for series convergence is that $a_n$ must be a zero sequence. A sufficient condition for series convergence is that its partial sum must also converge. My question is, is this condition also necessary? Are there convergent series whose partial sum isn't convergent?

Answer 1

In general, you can have all kinds of conditions. For example, consider the following candidates to be the "condition" that an integer $n$ is even:

Sufficient but not necessary: $n$ is divisible by $4.$

Necessary but not sufficient: $n \neq 1.$

Necessary and sufficient: The last digit of the decimal representation of $n$ is $0, 2, 4, 6,$ or $8.$

Neither necessary nor sufficient: $n$ is divisible by $3.$

---

Your example, however, is not just any mathematical statement. We usually define a convergent series by saying something like this:

The infinite series $\sum{a_n}$ is convergent if its sequence of partial sums converges.

(See here for example.)

Because this is a definition, by definition the condition "partial sums converge" is sufficient (the definition literally says so), and it is also necessary, because by definition that property, and not any other possible property, is what makes a series convergent.

Answer 2

Let $A$ be a proper subset of $B$ and let $B$ be a proper subset of $C$.

Then $x\in A$ is sufficient (but not necessary) for $x\in B$ and $x\in C$ is necessary (but not sufficient) for $x\in B$.

It is handsome to keep this picture/Venn diagram in mind.

Answer 3

The two qualifiers are used because they are independent.

For $x>1$, $x>0$ is necessary but not sufficient.

For $x>1$, $x>2$ is sufficient but not necessary.

For $x>1$, $y>1$ is neither necessary nor sufficient.

For $x>1$, $x-1>0$ is both necessary and sufficient.