Test for divergence

[MarkSA]

Hello,

I have questions about a couple cases of using the test for divergence.

First, I've seen the test for divergence used on:
The summation from n=1 to infinity of: [(-1)^n] * [sqrt(n)]
limit as n->infinity of [(-1)^n] * [sqrt(n)] = DNE (Does not exist) so divergent by the test
I assumed this is true because the (-1)^n bit oscillates and the limit of it does not exist, making it multiplied by the sqrt(n) irrelevant.

But if the above logic is true, wouldn't ANY alternating series automatically be divergent since taking the limit of the (-1)^(n-1) or similar terms, it would never exist due to oscillation? Can you explain?

Thanks

 

[pka]

There is a real difference is a null sequence, a sequences with limit zero, and others.
If \(\displaystyle \left( {a_n } \right) \to 0\; \Rightarrow \;\left( { - 1} \right)^n \left( {a_n } \right) \to 0\).
\(\displaystyle \frac{{\left( { - 1} \right)^n }}{{\sqrt n }} \to 0\;\& \;\left( { - 1} \right)^n \sqrt n (DNC)\), therefore \(\displaystyle \sum {\frac{{\left( { - 1} \right)^n }}{{\sqrt n }}} (conv)\& \sum {\;\left( { - 1} \right)^n \sqrt n } (DNC)\)

Lookup the exact wording of the alternating series test.

 

[MarkSA]

Ok, as I understand you, the series I posted can be found to be divergent by the Divergence test because a(sub n) goes to infinity. But if a(sub n) goes to 0, the test could not be used.

So the (-1)^n is simply ignored when using the divergence test? If I was showing work for it, would I need to include absolute value bars to cancel out the (-1)^n oscillating bit? Or can I simply ignore that anytime I do the divergence test and just do lim as n-> infinity of a(sub n) (where a(Sub n) is everything but the (-1) bit)

 

[pka]

The Divergence Test states that \(\displaystyle \left( {a_n } \right}) \not \to 0 \Rightarrow \quad \sum {a_n } \mbox{ diverges}\)
(i.e. if the sequence is not null the series diverges.)
There is nothing about \(\displaystyle (-1)^n\) in the statement.

 

[MarkSA]

And since all that (-1)^n does is change the sign, if the rest of it does not go to zero, then the series does not go to 0... correct? So (-1)^n can be ignored.

Ok, I think I understand now....