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**White_Owl****Member**- Registered: 2010-03-03
- Posts: 99

The problem is:

For which positive integers *k* is the following series convergent?

My Answer:

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...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi;

Something is wrong right there.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**White_Owl****Member**- Registered: 2010-03-03
- Posts: 99

I do not think there are mistakes:

Or are you talking about different equations?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Your first line is good.

Shouldn't manipulations maintain equality with the original assertion?

That does not?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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))

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

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.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**White_Owl****Member**- Registered: 2010-03-03
- Posts: 99

anonimnystefy, thank you. I see the mistake now

So my new answer is:

Therefore, series converges for k>=2

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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".

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**White_Owl****Member**- Registered: 2010-03-03
- Posts: 99

Yes, of course. Thank you.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Everything else seems okay to me.

You are welcome.

Here lies the reader who will never open this book. He is forever dead.

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