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## Topic review (newest first)

anonimnystefy
2013-04-01 14:44:27

Everything else seems okay to me.

You are welcome.

White_Owl
2013-04-01 14:39:54

Yes, of course. Thank you.

anonimnystefy
2013-04-01 10:32:10

Third line - in the denominator you have k1*k2*...*kn and you say below it "n times". It should be 1*2*...*kn and below it should be "kn times".

White_Owl
2013-04-01 10:28:04

anonimnystefy, thank you. I see the mistake now

So my new answer is:

Therefore, series converges for k>=2

anonimnystefy
2013-03-31 13:17:31

Well, the rest of his current work is okay. But that error is messing up the whole thing.

bobbym
2013-03-31 13:14:01

Hi;

That is what I mean, something is bad where I indicated. There could be further mistakes but that is where the first one occurs.

anonimnystefy
2013-03-31 13:09:23

#### White_Owl wrote:

I do not think there are mistakes:

Or are you talking about different equations?

The second line is not correc(k(n+1))!=1*2*3*...*(k(n+1)-1)*(k(n+1))

bobbym
2013-03-31 13:02:41

Your first line is good.

Shouldn't manipulations maintain equality with the original assertion?

That does not?

White_Owl
2013-03-31 12:54:10

I do not think there are mistakes:

Or are you talking about different equations?

bobbym
2013-03-31 12:10:38

Hi;

Something is wrong right there.

White_Owl
2013-03-31 11:31:44

The problem is:
For which positive integers k is the following series convergent?

For series to be convergent the next inequality should be true (by the Ratio Test):

Since we know that both k and n are positive we can omit absolute bars.

And now I simplify:

But since k is a constant this limit will never be less than 1. Therefore the series divergent for all possible k.

Did I make a mistake somewhere? Textbook is looking for a convergent series...