Series test for convergence or divergence

[Duane]

I had a bit of trouble in testing series like this for convergence
$$\sum_{ n=1 }^{ \infty } \frac { 1 }{ 2n+1 } $$
If by the comparison test, $\frac{ 1 }{ 2n+1 } < \frac{ 1 }{ 2n }$ for all of n>0,
and $\lim_{ n \rightarrow \infty} \frac{ 1 }{ 2n }$ =0, then the series should be convergent.
However, the correct answer in my book is divergent.
Why is that so?

Thanks for any help

PS. Could anyone help me with the LaTeX code? I can't seem to get it right.

 

[DonAntonio]

I had a bit of trouble in testing series like this for convergence
[tеx]\sum_{n=1}^{\infty}\frac{1}{2n+1}[/tеx]If by the comparison test, [tеx]\frac{1}{2n+1} < \frac{1}{2n}[/tеx] for all of n>0,
and [tеx]\lim_{n\rightarrow \infty}\frac {1}{2n}=0[/tеx], then the series should be convergent.
However, the correct answer in my book is divergent.
Why is that so?

Thanks for any help

PS. Could anyone help me with the LaTeX code? I can't seem to get it right.

I tried to edit your post but for some reason it keeps on coming up messed down...

Anyway, the comparison test works this time THE OTHER way:

\frac{1}{2n+1}\geq\frac{1}{2n+2n}=\frac{1}{4}\frac{1}{n} and since the rightmost sequence gives us the harmonic series times one fourth, it is divergent and thus is our series.

DonAntonio

 

[algebrat]

Yes, Don Antonio nailed it, but how should we recognize whether a series is convergent and divergent? Sometimes you can see it a little faster by recognizing the limiting behavior of the summands. For n large, you're summands will head toward \frac{1}{2n}. This could suggest to us quickly that it might diverge. (As always, still have to check by proving.)

Also, the problem with your proof. Pet P be the statement that a series converges, and let Q be the statement that the summands go to zero. So yes, P implies Q. But Q does not imply P, there are counterexamples, such as the harmonic series, which you discovered in a frustrating way.