# Thread: Does convergence of an infinite series imply convergence of the infinite sequence?

1. ## Does convergence of an infinite series imply convergence of the infinite sequence?

is it right to say something like,

"because the infinite series converges, the infinite sequence also converges"?

thanks!
explanations would be helpful if u wish to put some! =)

is it right to say something like,

"because the infinite series converges, the infinite sequence also converges"?

thanks!
explanations would be helpful if u wish to put some! =)

If you mean "because the inf. series SUM{a_n} converges then the inf. sequence {a_n} converges" then yes, you can say that....but it is a huge understatement, since if the series converges then not only the sequence converges but in fact it MUST converge to zero.
The other way around is false: a seq. may converge to zero but the inf. series formed with it can easily diverge.

Tonio

is it right to say something like,
"because the infinite series converges, the infinite sequence also converges"?
That statement is a bit vague. However here is a theorem.
The series $\sum\limits_{k = 1}^\infty {a_k }$ converges only if $\left( {a_n } \right) \to 0$.

Is that what you mean by the above statement?
This is often called first test for divergence.
You see that if $\left( {b_n }\right) \not\to 0$ then $\sum\limits_{k = 1}^\infty {b_k }$ must diverge.

On the other hand be careful. $\left( {\frac{1}{n}} \right) \to 0$ BUT $\sum\limits_{k = 1}^\infty {\frac{1}{k}}$ diverges.

4. Originally Posted by Plato
That statement is a bit vague. However here is a theorem.
The series $\sum\limits_{k = 1}^\infty {a_k }$ converges only if $\left( {a_n } \right) \to 0$.

No, the other way around me believes: If the series converge then a_n --> 0. The word "only" before the "if" doesn't change, I think, the direction of the logical implication.

Tonio

Is that what you mean by the above statement?
This is often called first test for divergence.
You see that if $\left( {b_n }\right) \not\to 0$ then $\sum\limits_{k = 1}^\infty {b_k }$ must diverge.

On the other hand be careful. $\left( {\frac{1}{n}} \right) \to 0$ BUT $\sum\limits_{k = 1}^\infty {\frac{1}{k}}$ diverges.
....

5. Originally Posted by tonio
No, the other way around me believes: If the series converge then a_n --> 0. The word "only" before the "if" doesn't change, I think, the direction of the logical implication.
Tonio
Tonio, How much do you know about implications?
It is well known that “P implies Q” is logically equivalent to “P only if Q”
These two statements are logically equivalent.
The series $\sum\limits_{k = 1}^\infty {a_k }$ converges only if $\left( {a_n } \right) \to 0$.

If the series $\sum\limits_{k = 1}^\infty {a_k }$ converges then $\left( {a_n } \right) \to 0$

6. Originally Posted by Plato
Tonio, How much do you know about implications?

I know quite a bit, thanx for asking.

It is well known that “P implies Q” is logically equivalent to “P only if Q”

I may be misunderstanding the language, but in my book , and until showed and convinced otherwise, the phrase "P only if Q" is the same as Q --> P or, in ordinary language, it means " P happens only if Q happens", and that's why I wrote that imo the word "only" does not change the direction of the implication.
The only way I can logically understand the word "only" wrt impication is in the notorious iff = if and only if, so "A iff and only if B" means "If B then A, and if A then B".

What I know is that P --> Q is logically eq. with ~ Q --> ~ P, with ~ meaning negation.

So unless you can convince me that in english language "P only if Q"
is the same as P --> Q I can't see how from the plain meaning of words we can reach your conclusion.

If you have some source where this is explained I'd thank you very much since, of course, I could be wrong and I could be misunderstanding some well-founded common understanding in logical mathematic spoken english which, btw, is my
3rd. langauge.

Tonio

These two statements are logically equivalent.
The series $\sum\limits_{k = 1}^\infty {a_k }$ converges only if $\left( {a_n } \right) \to 0$.

If the series $\sum\limits_{k = 1}^\infty {a_k }$ converges then $\left( {a_n } \right) \to 0$
.........................

7. Originally Posted by tonio
What I know is that P --> Q is logically eq. with ~ Q --> ~ P, with ~ meaning negation.
So unless you can convince me that in english language "P only if Q" is the same as P --> Q I can't see how from the plain meaning of words we can reach your conclusion.
If you have some source where this is explained I'd thank you very much since, of course, I could be wrong and I could be misunderstanding some well-founded common understanding in logical mathematic spoken english which, btw, is my 3rd. langauge.
If you access to any elementary logic textbook, in English, you will find a discussion of this topic.
There is nice page and a half discussion in Symbolic Logic by Irving M Copi
It is also discussed in a Set Theory Text by Stoll (ch. 4)
You can find it in almost any Discrete Mathematics textbook, say; Goodaire, Johnsonbaugh, Ross/Wright, or Grimaldi.

8. Originally Posted by Plato
If you access to any elementary logic textbook, in English, you will find a discussion of this topic.
There is nice page and a half discussion in Symbolic Logic by Irving M Copi
It is also discussed in a Set Theory Text by Stoll (ch. 4)
You can find it in almost any Discrete Mathematics textbook, say; Goodaire, Johnsonbaugh, Ross/Wright, or Grimaldi.

Yes, you were right. It appears in Suppes' "Introduction to logic", where it is explicitily stated that P --> Q can be expressed as "P only if Q".
There's also a nice discussion in No. 10 at Peter Suber, "Translation Tips" which seems to explain my confussion with spoken language. I also calles the "only if" the least intuitive (logical connectives translation) because, apparently, of the language bias we have.
And this is a nice a day since I learned something new.
Thanx

Tonio