# Convergent -divergent series

1. Nov 30, 2009

### 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??

2. Nov 30, 2009

### 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?

Last edited: Nov 30, 2009
3. Nov 30, 2009

You put it in using an $tag. 4. Dec 1, 2009 ### evagelos 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 5. Dec 1, 2009 ### g_edgar Contrapositive? 6. Dec 1, 2009 ### fourier jr if $$b_{n} = (1+1/n)a_{n}$$ then they're equivalent. they're converses of each other 7. Dec 1, 2009 ### 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 8. Dec 1, 2009 ### 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. Last edited: Dec 1, 2009 9. Dec 2, 2009 ### HallsofIvy The statement, "if the series does not diverge to infinity it means that the series converges" is true for positive series. The series [itex]\sum_{n=0}^\infty (-1)^n$ neither diverges to infinity nor converges.

10. Dec 2, 2009

### evagelos

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 ??

Last edited: Dec 2, 2009
11. Dec 2, 2009

### 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

12. Dec 2, 2009

### evagelos

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

13. Dec 2, 2009

### 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 $|b_n| \leq |a_n|$

14. Dec 2, 2009

### evagelos

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

15. Dec 2, 2009

### fourier jr

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

16. Dec 3, 2009

### evagelos

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??

17. Dec 3, 2009

### 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"

18. Dec 3, 2009

### 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 :$$n\geq N\Longrightarrow$$ $\sum_{k=1}^{n}a_{k} \geq\epsilon$ ??

19. Dec 3, 2009

### 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.

20. Dec 4, 2009

### evagelos

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

21. Dec 4, 2009

### 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.

Last edited: Dec 4, 2009
22. Dec 4, 2009

### 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.

23. Dec 4, 2009

### 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?

24. Dec 4, 2009

### D H

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

25. Dec 4, 2009

### 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,

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

Last edited: Dec 4, 2009