Do Divergent Series Always Imply Logical Equivalences in Convergence Statements?

[evagelos]

Suppose the series \sum a_{n} diverges to +\infty,

Then if the series does not diverge to infinity it means that the series converges, and

consequently the statement : if \sum a_{n} diverges ,then \sum b_{n} diverges,is equivalent to :

if \sum b_{n} converges ,then \sum a_{n} converges??

 

[fourier jr]

You need an additional condition on the b_n; if \sum a_{n} diverges and |b_n| \geq |a_n| for all n then \sum b_{n} diverges also. & conversely, if \sum a_{n} converges and |b_n| \leq |a_n| for all n then \sum b_{n} converges. (it's called the comparison test)

ps- to anybody else who knows, how do I put the latex in line?

 

[n!kofeyn]

You put it in using an tag.

 

[evagelos]

You need an additional condition on the b_n; if \sum a_{n} diverges and |b_n| \geq |a_n| for all n then \sum b_{n} diverges also. & conversely, if \sum a_{n} converges and |b_n| \leq |a_n| for all n then \sum b_{n} converges. (it's called the comparison test)

ps- to anybody else who knows, how do I put the latex in line?

So if we say that :if \sum a_{n} diverges, then prove that ,\sum(1+1/n)a_{n} diverges ,

it is not equivalent to :

if \sum(1+1/n)a_{n} converges ,then \sum a_{n} converges

 

[g_edgar]

So if we say that :if \sum a_{n} diverges, then prove that ,\sum(1+1/n)a_{n} diverges ,

it is not equivalent to :

if \sum(1+1/n)a_{n} converges ,then \sum a_{n} converges

Contrapositive?

 

[fourier jr]

So if we say that :if \sum a_{n} diverges, then prove that ,\sum(1+1/n)a_{n} diverges ,

it is not equivalent to :

if \sum(1+1/n)a_{n} converges ,then \sum a_{n} converges

if b_{n} = (1+1/n)a_{n} then they're equivalent. they're converses of each other

 

[evagelos]

Assuming the series \sum a_{n} diverges to +\infty, then we have to show that the series \sum(1+1/n)a_{n} diverges to +\infty.

But according to contrapositive law it is equivalent to show:

if \sum(1+1/n)a_{n} does not diverge to +\infty ,then \sum a_n} does not diverge to +\infty

is that O.K

 

[fourier jr]

that's right, although people usually just say converge rather than "does not diverge" even though they mean the same thing. & if this is for a homework problem you should probably mention the comparison test somewhere.

 

[HallsofIvy]

The statement, "if the series does not diverge to infinity it means that the series converges" is true for positive series. The series \sum_{n=0}^\infty (-1)^n neither diverges to infinity nor converges.

 

[evagelos]

that's right, although people usually just say converge rather than "does not diverge" even though they mean the same thing. & if this is for a homework problem you should probably mention the comparison test somewhere.

But,

when we say that a series \sum b_{n} does not converge to +\infty it does not mean that the series converges to a limit b,

because

When \sum b_{n} diverges to +\infty ,by definition we have:

for all ε>0 there exists a natural No N such that for all ,n\geq N\Longrightarrow \sum_{k=1}^{n}b_{k}\geq\epsilon

and consequently,

if the series does not diverge to infinity we have:

there exists an ε>0 and for all natural Nos N there exists an ,n\geq N and \sum_{k=1}^{n}b_{k}<\epsilon.

Does that mean that \sum b_{n} converges to b ??

 

[fourier jr]

let me make sure i have my facts straight (& practicing my itex-messaging...)

- a series converges to A if for all \epsilon > 0 there is an N with the property that for k>N, |(\sum_{n=1}^{k}a_{n}) - A| < \epsilon

- & if it doesn't converge it diverges, either to infinity or it alternates forever like (-1)^n because there's no N with the above property (thx to halls for reminding me )

- the comparison test says that if \sum_{n=1}^{\infty}a_n converges, and |b_n| \leq |a_n| for all n, then \sum_{n=1}^{\infty}b_n converges. or conversely, if \sum_{n=1}^{\infty}a_n diverges, and |b_n| \geq |a_n| for all n, then \sum_{n=1}^{\infty}b_n diverges.

- in particular (set b_n = (1+1/n)a_n), if given that \sum_{n=1}^{\infty}(1+1/n)a_n converges, then \sum_{n=1}^{\infty}a_n converges by comparison with \sum_{n=1}^{\infty}(1+1/n)a_n, since |(1+1/n)a_n| \geq |a_n| for all n

- or looking at it the other way, if \sum_{n=1}^{\infty}a_n diverges, then \sum_{n=1}^{\infty}(1+1/n)a_n diverges by comparison with \sum_{n=1}^{\infty}a_n since |(1+1/n)a_n| \geq |a_n| for all n

 

[evagelos]

let me make sure i have my facts straight (& practicing my itex-messaging...)

- a series converges to A if for all \epsilon > 0 there is an N with the property that for k>N, |(\sum_{n=1}^{k}a_{n}) - A| < \epsilon

- & if it doesn't converge it diverges, either to infinity or it alternates forever like (-1)^n because there's no N with the above property (thx to halls for reminding me )

- the comparison test says that if \sum_{n=1}^{\infty}a_n converges, and |b_n| \leq |a_n| for all n, then \sum_{n=1}^{\infty}b_n converges. or conversely, if \sum_{n=1}^{\infty}a_n diverges, and |b_n| \geq |a_n| for all n, then \sum_{n=1}^{\infty}b_n diverges.

- in particular (set b_n = (1+1/n)a_n), if given that \sum_{n=1}^{\infty}(1+1/n)a_n converges, then \sum_{n=1}^{\infty}a_n converges by comparison with \sum_{n=1}^{\infty}(1+1/n)a_n, since |(1+1/n)a_n| \geq |a_n| for all n

- or looking at it the other way, if \sum_{n=1}^{\infty}a_n diverges, then \sum_{n=1}^{\infty}(1+1/n)a_n diverges by comparison with \sum_{n=1}^{\infty}a_n since |(1+1/n)a_n| \geq |a_n| for all n

Nowhere is given that \sum_{n=1}^{\infty}(1+1/n)a_n converges.

And i think that the converse of divergence to infinity, is not convergence ,as i tried to show in my post # 10

 

[fourier jr]

a series converges (to a number A) if for all \epsilon > 0 there is an N with the property that for k>N, |(\sum_{n=1}^{k}a_{n}) - A| < \epsilon (not what you wrote in post #10) & it diverges if there is no N. The opposite of diverge is converge, which means there is such an N for all epsilon. It doesn't matter whether it diverges to infinity or oscillates forever. & if, as in the original post, you want to use one series to determine convergence or divergence of another there needs to be a way of comparing them, and the usual way, at least if you want to use the comparison test it's whether or not |b_n| \leq |a_n|

 

[evagelos]

a series converges (to a number A) if for all \epsilon > 0 there is an N with the property that for k>N, |(\sum_{n=1}^{k}a_{n}) - A| < \epsilon (not what you wrote in post #10) & it diverges if there is no N.

Do you agree that the negation of the above definition implies divergence of the series ?

 

[fourier jr]

that's what I said. It diverges if there's no N.

 

[evagelos]

that's what I said. It diverges if there's no N.

But the negation of the definition of convergence is the following :

For all ,A there exists ε>0 and for all N there exists k>N and
|(\sum_{n=1}^{k}a_{n}) - A| \geq\epsilon.

Do you agree that this denotes the divergence of the series??

 

[fourier jr]

I think that looks pretty close. I would change the "there exists k>N" to something like "when k>N" or "for all k>N"

 

[evagelos]

Would you agree then that the denial to the divergence of a series to +\infty is :

There exists an ε>0 and for all natural Nos N there exists an n\geq N and \sum_{k=1}^{n}a_{k}) <\epsilon

Provided the definition of a series diverging to +\infty is:

for all ε>0 there exists a natural No N and for all ,n :n\geq N\Longrightarrow \sum_{k=1}^{n}a_{k} \geq\epsilon ??

 

[fourier jr]

the above "definition" of divergence doesn't necessarily mean it goes to infinity, it means that the partial sums doesn't get closer to that number A.

 

[evagelos]

How would you define a series that go to +\infty

 

[fourier jr]

If you remember, the sum of an infinite series is the limit of the sequence of partial sums, in other words, \sum_{k=0}^{\infty}a_k = \lim_{n \rightarrow \infty}\sum_{k=0}^{n}a_k. The series diverges if limit doesn't exist, or is infinite. Sorry for the confusion, I should have mentioned it a long time ago.

 

[evagelos]

K.G.BINMORE in his book Mathematical Analysis ,pages 38 ,39 gives a definition for a sequence diverging to +\infty,which is the following:

" We say that a sequence x_{n} diverges to +\infty and write x_{n}\rightarrow +\infty as n\rightarrow +\infty if, for any H>0,we can find an N such that,for any n>N, x_{n}>H"

Now if X_{n} denotes the partial sums of a series ,then for a series to diverge to +\infty the above definition is applicable.

That definition i have repeatedly written in my previous posts ,but you have not accepted.

 

[fourier jr]

i had never heard of that before. the definition i learned was just that it didn't converge, and that the limit didn't exist. if you know what you're talking about why are you asking then?

 

[D H]

But,

when we say that a series \sum b_{n} does not converge to +\infty it does not mean that the series converges to a limit b,

because

When \sum b_{n} diverges to +\infty ,by definition we have:

for all ε>0 there exists a natural No N such that for all ,n\geq N\Longrightarrow \sum_{k=1}^{n}b_{k}\geq\epsilon

and consequently,

if the series does not diverge to infinity we have:

there exists an ε>0 and for all natural Nos N there exists an ,n\geq N and \sum_{k=1}^{n}b_{k}<\epsilon.

Does that mean that \sum b_{n} converges to b ??

You cannot infer that. Halls already showed a series in post #9 that neither diverges to infinity nor converges to a limit.

 

[hamster143]

This whole discussion baffles me. I haven't seen so many undefined variables, lapses of logic, mismatched brackets and meaningless statements in one place for a long time.

What exactly does this mean, for example,

So if we say that "if A then prove that B"

it is not equivalent to "if C then D"

I have a vague feeling that I'm asked to prove something, but I'm not sure what...

 

[evagelos]

Why use words you do not understand. What are "lapses of logic"