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"