If diverges

show that:

diverges (Wink)

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- Nov 26th 2009, 10:26 PMXingyuanWho come to beat this ：）
If diverges

show that:

diverges (Wink) - Nov 26th 2009, 10:33 PMRockHard
I never get these general term series question, multiplied by some function

I know that simply by the divergence test or that it is the harmonic series.

which is not 0 and by the divergence test if the limit of diverges

Could say that the An is just like multiplying a constant and which multiplying by a constant will not affect if a series diverges or convergences? just the sum?

similarly

This is a Harmonic series which diverges...I am incorrect? - Nov 26th 2009, 10:49 PMDefunkt
.

What you have here is either a sum of two divergent series or a sum of a convergent and a divergent series, both of which diverge. - Nov 26th 2009, 11:01 PMRockHard
Defunkt, would the way I TRIED, lol, to show this be applicable? more specifically the last bit of my reply

- Nov 26th 2009, 11:05 PMDefunkt
Well, I don't really understand what you're saying there :P Try organizing your thoughts more clearly and then presenting them.

If you are saying that multiplying the series by is like multiplying by a constant, that's wrong. It simply is not a constant. And no, it is not a harmonic series -- http://en.wikipedia.org/wiki/Harmoni...s_(mathematics) this is the only form of a harmonic series. - Nov 26th 2009, 11:10 PMRockHard
Thanks for the tips, :p I am just so eager to learn, I make a mess of myself, I was trying to say that multiplying by to the series of

, you can treat as constant, more than likely this is where I am incorrect, and that multiplying a constant to a series will not affect its convergence or divergence just it's sum.

Also tried saying that

can be written as

which looks to be of similar form to the harmonic series to me, but what the hell, I am more incorrect than correct, when I post messes like these, lol. - Nov 26th 2009, 11:16 PMDefunkt
The harmonic series is ONLY this: . NOTHING else. Not everything that is divided by n is a harmonic series :P

The part about multiplying by a constant is true, though. If you multiply a series by a non-zero constant, its convergence will not change -- only the value of the series (if it converges). The thing here is that is not a constant (or do you mean the original divergent series ? either way, both aren't constants), even though it goes to 1 as n goes to infinity. - Nov 26th 2009, 11:23 PMRockHard
Thanks for the confirmation, so this is best shown using the property of adding two series together will only converge if both series are convergent, correct?

- Nov 26th 2009, 11:42 PMXingyuan
- Nov 26th 2009, 11:52 PMShanks
let

then .

Since divergens, then at least one of them diverges.

If only one of them diverges; then obviously the conclusion is true.

If both of them diverges, thennnnnnn......

what happens if diverges to ? - Nov 26th 2009, 11:55 PMRockHard
Well, I think that if the way Defunkt looked at the function as a sum of two series, and the property is for that too be convergent both series he broke apart must be convergent

- Nov 26th 2009, 11:58 PMShanks
maybe, we can think in a inverse direction, maybe it is easier to prove:

If converges,

then converges. - Nov 27th 2009, 12:15 AMDefunkt
Err.. yes, my mistake. It would probably be easier to prove noting that

1) diverges

2) - Nov 27th 2009, 12:39 AMXingyuan
- Nov 27th 2009, 01:55 AMDefunkt
lol, sorry again. shouldn't post so early in the morning. the equality will hold in absolute value, and that should work... off i go now :S