When Is the Divergence Test Applicable for Series?

steel1
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Homework Statement


Not really a problem, more of a general question. When exactly can you use the Divergence test. Does it only work on both series and sequences?

Homework Equations


The series Diverges if lim ƩAn ≠ 0

The Attempt at a Solution


If you take the lim of the series n^3/2n^3 ≠ 0 there it diverges.

Now, look at the series (n+1)/n(n+2). You have to use the integral test to show convergence or divergence for this. After doing it, you get the series Diverges. Why can't i just use l'hospitals rule on the 2nd series, and get 1/2n, then take the limit. And it should converge to zero.

Is it because i used l'hospitals rule therefore i can not use the divergence test anymore?
 
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steel1 said:

Homework Statement


Not really a problem, more of a general question. When exactly can you use the Divergence test. Does it only work on both series and sequences?


Homework Equations


The series Diverges if lim ƩAn ≠ 0
NO, the series diverges if ##\lim_{n\to\infty}A_n\ne 0## (no sum).

The Attempt at a Solution


If you take the lim of the series n^3/2n^3 ≠ 0 there it diverges.

Now, look at the series (n+1)/n(n+2). You have to use the integral test to show convergence or divergence for this. After doing it, you get the series Diverges. Why can't i just use l'hospitals rule on the 2nd series, and get 1/2n, then take the limit. And it should converge to zero.

Is it because i used l'hospitals rule therefore i can not use the divergence test anymore?

If the nth term doesn't go to zero the series diverges. But the nth term may go to zero and yet the series diverges anyway. So the test for divergence is sufficient but not necessary for divergence. That's why you need other more sophisticated tests.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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